Category: ML Aggarwal Solutions

The availability of step-by-step ML Aggarwal Class 12 Solutions Chapter 8 Integrals Ex 8.12 can make challenging problems more manageable.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.12

Very Short answer type questions (1 to 5) :

Evaluate the following (1 to 12) integrals

Question 1.
(i) ∫ e^x (sin x + cos x) dx (NCERT)
(ii) ∫ e^x (tan x + sec² x) dx
Solution:
(i) ∫ e^x (sin x + cos x) dx
= ∫ e^x sin x dx + ∫ e^x cos x dx
= sin x e^x – ∫ cos x e^x dx + ∫ e^x cos x dx + C
= sin x . e^x + C

(ii) Let I = ∫ e^x (tan x + sec² x) dx
= ∫ e^x tan x dx + ∫ e^x sec² x dx
= tan x . e^x – ∫ sec² x . e^x dx + ∫ e^x . sec² x dx + C
= tan x . e^x + C

Question 2.
(i) ∫ e^x (log x + \(\frac{1}{x}\)) dx
(ii) ∫ e^x (cot x – cosec² x) dx
Solution:
(i) Let I = ∫ e^x (log x + \(\frac{1}{x}\)) dx
= ∫ e^x log x dx + ∫ \(\frac{e^x}{x}\) dx
= log x e^x – ∫ \(\frac{1}{x}\) e^x dx + ∫ \(\frac{e^x}{x}\) dx + c
= e^x log x + c

(ii) Let I = ∫ e^x (cot x – cosec² x) dx
= ∫ e^x [cot x + (- cosec² x)] dx
= ∫ e^x cot x dx – ∫ e^x cosec² x dx
= e^x cot x + ∫ cosec² x e^x dx – ∫ e^x cosec² x dx
= e^x cot x + c

Question 3.
(i) ∫ e^x (x² + 2x) dx
(ii) ∫ (x + 1) e^x dx
Solution:
(i) Let I = ∫ e^x (x² + 2x) dx
= ∫ e^x x² dx + ∫ e^x . 2x dx
= x² e^x – 2x e^x dx + 2x . e^x dx + C
= x² e^x + C’

(ii) Let I = ∫ (x + 1) e^x dx
= ∫ x e^x dx + ∫ e^x dx
= x e^x – ∫ 1 . e^x dx + ∫ e^x + C
= x e^x + C

Question 4.
(i) ∫ e^x \(\left(\frac{1}{x}-\frac{1}{x^2}\right)\) dx
(ii) ∫ (1 + log x) dx
Solution:
(i) Let I = ∫ e^x \(\left(\frac{1}{x}-\frac{1}{x^2}\right)\) dx

(ii) Let I = ∫ (1 + log x) dx
= ∫ 1 . dx + ∫ log x . 1 dx
= x + log x . x – ∫ \(\frac{1}{x}\) . x dx
= x + x log x – x + C
= x log x + C

Question 5.
(i) ∫ e^x (sin^-1 x + \(\frac{1}{\sqrt{1-x^2}}\)) dx
(ii) ∫ (x cos x + sin x) dx
Solution:
(i) Let I = ∫ e^x (sin^-1 x + \(\frac{1}{\sqrt{1-x^2}}\)) dx
= ∫ e^x sin^-1 x dx + ∫ \(\frac{e^x}{\sqrt{1-x^2}}\) dx
= e^x sin^-1 x – ∫ \(\frac{1}{\sqrt{1-x^2}}\) e^x dx + ∫ \(\frac{e^x d x}{\sqrt{1-x^2}}\) + c
= e^x sin^-1 x + c

(ii) Let I = ∫ (x cos x + sin x) dx + ∫ sin x dx
= x sin x – ∫ sin x dx + ∫ sin x dx + C
= x sin x + C

Question 6.
(i) Given ∫ e^x (tan x + 1) sec x dx = e^x f(x) + C. Write f(x) satisfying the above.
(ii) If ∫ \(\left(\frac{x-1}{x^2}\right)\) e^x dx = f(x) e^x + C, then write the value of f(x).
Solution:
(i) Let I = ∫ e^x (tan x + 1) sec x dx
= ∫ e^x tan x sec x dx + ∫ e^x sec x dx
= e^x secx – ∫ e^x sec x dx + ∫ e^x sec x dx + C
= e^x sec x + C ……………..(1)
Also I = e^x f(x) + c ……………….(2)
From (1) and (2) ; we have
f(x) = sec x

(ii) Let I = ∫ \(\left(\frac{x-1}{x^2}\right)\) e^x dx

Also given I = f(x) e^x + C …………(2)
From (1) and (2) ; we have
f(x) = \(\frac{1}{x}\)

Question 7.
(i) ∫ e^x (tan x + log sec x) dx
(ii) ∫ \(\frac{2 x-1}{4 x^2}\) e^2x dx
Solution:
(i) Let I = ∫ e^x (tan x + log sec x) dx
= ∫ e^x [tan x + (- log cos x)] dx
= – [ (log cos x) e^x – ∫ \(\frac{1}{cos x}\) (- sin x) e^x dx] + ∫ e^x tan x dx
= – e^x log cos x + c
= e^x log (cos x)^-1 + c
= e^x log sec x + c

(ii) Let I = ∫ \(\frac{2 x-1}{4 x^2}\) e^2x dx

Question 8.
(i) ∫ \(\frac{\sin x \cos x-1}{\sin ^2 x}\) e^x dx
(ii) ∫ e^x (cot x + log |sin x|) dx
Solution:
(i) Let I = ∫ \(\frac{\sin x \cos x-1}{\sin ^2 x}\) e^x dx
= ∫ e^x [cot x – cosec² x] dx
= ∫ e^x cot x – ∫ e^x cosec² x dx
= cot x e^x – ∫ (- cosec² x) e^x dx – ∫ e^x cosec² x dx
= e^x cot x + ∫ e^x cosec² x – ∫ e^x cosec² x dx + c
= e^x cot x + c

(ii) Let I = ∫ e^x (cot x + log |sin x|) dx
= ∫ e^x log sin x dx + ∫ e^x cot x dx
= (log sin x) e^x – ∫ \(\frac{1}{sin x}\) cos x e^x dx + ∫ e^x cot x dx
= e^x log sin x – ∫ cot x e^x dx + ∫ e^x cot x dx + C
= e^x log sin x + c

Question 9.
(i) ∫ \(\frac{x e^x}{(x+1)^2}\) dx (NCERT)
(ii) ∫ \(\frac{x-1}{(x+1)^3}\) e^x dx
Solution:
(i) Let I = ∫ \(\frac{x e^x}{(x+1)^2}\) dx

(ii) Let I = ∫ e^x \(\frac{x-1}{(x+1)^3}\) dx

Question 10.
(i) ∫ \(\frac{x-3}{(x-1)^3}\) e^x dx (NCERT)
(ii) ∫ \(\frac{x e^{2 x}}{(1+2 x)^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{x-3}{(x-1)^3}\) e^x dx

(ii) Let I = ∫ \(\frac{x e^{2 x}}{(1+2 x)^2}\) dx

Question 11.
(i) ∫ \(\frac{1+\sin x}{1+\cos x}\) e^x dx
(ii) ∫ \(\frac{2+\sin 2 x}{1+\cos 2 x}\) e^x dx
Solution:
(i) Let I = ∫ \(\frac{1+\sin x}{1+\cos x}\) e^x dx

(ii) Let I = ∫ \(\frac{2+\sin 2 x}{1+\cos 2 x}\) e^x dx
= ∫ \(\left[\frac{2+2 \sin x \cos x}{2 \cos ^2 x}\right]\) e^x dx
= ∫ sec² x . e^x dx + ∫ e^x . tan x dx
= e^x tan x – ∫ e^x . tan x dx + ∫ e^x tan x dx + C
= e^x tan x + C

Question 12.
(i) ∫ \(\frac{\sin 4 x-4}{1-\cos 4 x}\) e^x dx
(ii) ∫ (cos x + 3 sin x) e^3x dx
Solution:
(i) Let I = ∫ e^x \(\frac{\sin 4 x-4}{1-\cos 4 x}\) dx
= ∫ e^x \(\left(\frac{\sin 4 x-4}{2 \sin ^2 2 x}\right)\) dx
= ∫ e^x \(\left[\frac{2 \sin 2 x \cos 2 x-4}{2 \sin ^2 d x}\right]\) dx
[Form ∫ e^x [f(x) + f'(x)] dx]
= ∫ e^x [cot 2x – 2 cosec² 2x] dx
∴ I = ∫ e^x cot 2x dx – 2 ∫ e^x cosec² 2x dx
= cot 2x e^x – ∫ – cosec² 2x . 2 e^x dx – 2 ∫ e^x cosec² 2x dx
= e^x . cot 2x + C

(ii) Let I = ∫ (cos x + 3 sin x) e^3x dx
= ∫ e^3x cos x dx + ∫ sin x e^3x dx
= e^3x sin x – ∫ 3 e^3x sin x dx + 3 ∫ sin x e^3x dx + C
= e^3x sin x + C

Question 12 (old).
(i) ∫ (1 + log x) dx
Solution:
(i) Let I = ∫ (1 + log x) dx
= ∫ log x . dx + ∫ dx
= x log x – ∫ \(\frac{1}{x}\) . x dx + ∫ dx + C
= x log x + C

Question 13.
∫ (sin (log x) + cos (log x)) dx
Solution:
Let I = ∫ (sin (log x) + cos (log x)) dx
put log x = t
⇒ x = e^t
⇒ dx = e^t dt
∴ I = ∫ [sin t + cos t] e^t dt
= ∫ e^t sin t dt + ∫ e^t cos t dt
= sin t e^t – ∫ cos t e^t dt + ∫ e^t cos t dt
= e^t sin t + C
= x sin (log x) + C

Students often turn to Class 12 ISC Maths Solutions Chapter 8 Integrals Ex 8.11 to clarify doubts and improve problem-solving skills.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.11

Very Short answer type questions (1 to 3) :

Evaluate the following (1 to 22) integrals

Question 1.
(i) ∫ x e^x dx
(ii) ∫ x sin x dx (NCERT)
Solution:
(i) ∫ x e^x dx = x . e^x – ∫ 1 . e^x dx
[Integrating by parts]
= x e^x – e^x + C
= (x – 1) e^x + C

(ii) ∫ x sin x dx = – x cos x – ∫ 1 . (- cos x) dx
= – x cos x + sin x + C

Question 2.
(i) ∫ x sec² x dx (NCERT)
(ii) ∫ x sin 3x dx (NCERT)
Solution:
(i) ∫ x sec² x dx
= x tan x – ∫ 1 . tan x dx
= x tan x – ∫ \(\frac{\sin x}{\cos x}\) dx
= x log x + log |cos x| + C

(ii) ∫ x sin 3x dx
= x \(\left(-\frac{\cos 3 x}{3}\right)\) + ∫ 1 . \(\frac{\cos 3 x}{3}\) dx
= – x \(\frac{\cos 3 x}{3}\) + \(\frac{\sin 3 x}{9}\) + C

Question 3.
(i) ∫ x e^2x dx
(ii) ∫ x cos 2x dx
Solution:
(i) ∫ x e^2x dx = \(\frac{x e^{2 x}}{2}-\int 1 \cdot \frac{e^{2 x}}{2} d x\)
= \(\frac{x e^{2 x}}{2}-\frac{e^{2 x}}{4}\)
= \(\frac{1}{4}\) (2x – 1) e^2x + C

(ii) ∫ x cos 2x dx = \(x \frac{\sin 2 x}{2}-\int 1 \cdot \frac{\sin 2 x}{2} d x\)
= \(\frac{x \sin 2 x}{2}+\frac{\cos 2 x}{4}\) + C

Question 4.
(i) ∫ x sec² x tan x dx
(ii) ∫ (e^{log x} + sin x) cos x dx
Solution:
(i) ∫ x sec² x tan x dx
= x . \(\frac{\tan ^2 x}{2}\) – ∫ 1 . \(\frac{\tan ^2 x}{2}\) dx
[∵ ∫ sec² x tan x dx = ∫ tan x . sec² x dx
= \(\frac{\tan ^2 x}{2}\) ;
Since ∫ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\), n ≠ – 1]
= \(\frac{x \tan ^2 x}{2}\) – \(\frac{1}{2}\) ∫ (sec² x – 1) dx
= \(\frac{x \tan ^2 x}{2}\) – \(\frac{1}{2}\) tan x + \(\frac{x}{2}\) + C
= \(\frac{1}{2}\) [x (1 + tan² x) – tan x] + C
= \(\frac{1}{2}\) [x sec² x – tan x] + C

(ii) Let I = ∫ (e^{log x} + sin x) cos x dx
= ∫ (x + sin x) cos x dx
= ∫ x cos x dx + ∫ \(\frac{1}{2}\) sin 2x dx
= x sin x – ∫ 1 . sin x dx – \(\frac{\cos 2 x}{4}\) + c
= x sin x + cos x – \(\frac{\cos 2 x}{4}\) + c

Question 5.
(i) ∫ x log x dx (NCERT)
(ii) ∫ x log 2x (NCERT)
Solution:
(i) Let I = ∫ x log x dx
using integrating by parts
∴ I = log x . \(\frac{x^2}{2}\) – ∫ \(\frac{1}{x} \cdot \frac{x^2}{2}\) dx
= \(\frac{x^2}{2}\) log x – \(\frac{x^2}{4}\) + C
= \(\frac{x^2}{4}\) (2 log x – 1) + C

(ii) ∫ x log 2x dx = log 2x . \(\frac{x^2}{2}\) – ∫ \(\frac{2}{2 x} \frac{x^2}{2}\) dx
[Integrating by parts]
= \(\frac{x^2}{2}\) log 2x – \(\frac{x^2}{4}\) + C
= \(\frac{x^2}{4}\) [2 log 2x – 1] + C

Question 6.
(i) ∫ x⁴ log x dx
(ii) ∫ log x dx
Solution:
(i) Let I = ∫ x⁴ log x dx
= log x . \(\frac{x^5}{5}\) – ∫ \(\frac{1}{x} \cdot \frac{x^5}{5}\) dx + C
= \(\frac{x^5}{5} \log x-\frac{1}{5} \cdot \frac{x^5}{5}\) + C
= \(\frac{x^5}{25}\) (5 log x – 1) + C

(ii) ∫ log x dx = ∫ log x . 1 dx
= x log x – ∫ \(\frac{1}{x}\) . x dx
= x log x – x + C
= x (log x – 1) + C

Question 6 (old).
(i) ∫ x² log x dx (NCERT)
Solution:
(i) ∫ x²
= \(\log x \cdot \frac{x^3}{3}-\int \frac{1}{x} \cdot \frac{x^3}{3} d x\)
= \(\frac{x^3}{3} \log x-\frac{x^3}{9}\) + C
= \(\frac{x^3}{9}\) (3 log x – 1) + C

Question 7.
(i) ∫ (x² + 1) log x dx (NCERT)
(ii) ∫ x² log (1 + x) dx
Solution:
(i) ∫ (x² + 1) log x dx
= log x . (\(\frac{x^3}{3}\) + x) – ∫ \(\frac{1}{x}\left(\frac{x^3}{3}+x\right)\)
[using integrating by parts]
∴ I = (\(\frac{x^3}{3}\) + x) log x – \(\frac{1}{3} \frac{x^3}{3}\) – x + C
= \(\frac{x^3}{9}\) (3 log x – 1) + x (log x – 1) + C

(ii) ∫ x² log (1 + x) dx

Question 8.
(i) ∫ tan^-1 x dx (NCERT)
(ii) ∫ cos^-1 \(\left(\frac{1}{x}\right)\) dx
Solution:
(i) ∫ tan^-1 x dx
= ∫ tan^-1 x . 1 dx
= x tan^-1 x – ∫ \(\frac{x}{1+x^2}\) dx
= x tan^-1 x – \(\frac{1}{2}\) log (1 + x²) + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]

(ii) ∫ cos^-1 \(\frac{1}{x}\) dx
= ∫ sec^-1 x dx
= ∫ sec^-1 x . 1 dx
= x sec^-1 x – ∫ \(\frac{1}{x \sqrt{x^2-1}}\) x dx
= x sec^-1 x – ∫ \(\frac{d x}{\sqrt{x^2-1}}\)
= x sec^-1 x – log |x + \(\sqrt{x^2-1}\) + C

Question 9.
(i) ∫ x³ tan^-1 x dx
(ii) ∫ x² sin^-1 x dx
Solution:
(i) Let I = ∫ x³ tan^-1 x dx

(ii) Let I = ∫ x² sin^-1 x dx

Question 10.
(i) ∫ \(\frac{\sin ^{-1} x}{\sqrt{1-x}}\) dx
(ii) ∫ x² e^x dx
Solution:
(i) Let I = ∫ \(\frac{\sin ^{-1} x}{\sqrt{1-x}}\) dx

(ii) ∫ x² e^x dx
= x² e^x – ∫ 2x . e^x dx [Integrating by parts]
= x² e^x – 2 [x e^x – ∫ 1 . e^x dx] [Integrating by parts]
= x² e^x – 2 [x e^x – e^x] + C
= e^x (x² – 2x + 2) + C

Question 11.
(i) ∫ x² e^{3 x} dx
(ii) ∫ x² cos x dx
Solution:
(i) ∫ x² e^{3 x} dx

(ii) ∫ x² cos x dx
= x² sin x – ∫ 2x sin x dx
= x² sin x – 2 [- x cos x + ∫ 1 . cos x dx]
= x² – 2 [- x cos x + ∫ 1 . cos x dx]
= x² – 2 [- x cos x + sin x] + c
= x² sin x + 2x cos x – 2 sin x + c

Question 12.
(i) ∫ (log x)² dx
(ii) ∫ x³ sin (x²) dx
Solution:
(i) ∫ (log x)² dx
= ∫ (log x)² . 1 dx
= (log x)² x – ∫ 2 log x . \(\frac{1}{x}\) . x dx
= x (log x)² – 2 ∫ log x . 1 dx
= x (log x)² – 2 [x log x – x] + C
= x (log x)² – 2x log x + 2x + C

(ii) Let I = ∫ x³ sin (x²) dx
put x² = t
⇒ 2x dx = dt
= ∫ t sin t \(\frac{dt}{2}\)
= \(\frac{1}{2}\) [- t cos t + ∫ 1 . cos t dt] + C
= \(\frac{1}{2}\) [- t cos t + sin t] + C
= \(\frac{1}{2}\) [- x² cos² x² + sin x²] + C

Question 13.
(i) ∫ 2x³ ex² dx
(ii) ∫ cos √x dx
Solution:
(i) Let I = ∫ 2x³ ex² dx
put x² = t
⇒ 2x dx = dt
= ∫ t . e^t dt
= t e^t – ∫ 1 . e^t dt + C
= t e^t – e^t + C
= (t – 1) e^t + C
= (x² – 1) e^x2 + C

(ii) Let I = ∫ cos √x + C
put √x = t
⇒ x = t²
⇒ dx = 2t dt
= ∫ cos t (2t dt)
= 2 ∫ t cos t dt
= 2 [t sin t – ∫ 1 . sin t dt] + c
= 2 [t sin t + cos t] + c
= 2 [√x sin √x + cos √x] + c

Question 14.
(i) ∫ tan^-1 √x dx
(ii) ∫ x³ tan^-1 (x²) dx
Solution:
(i) Let I = ∫ tan^-1 √x dx
put √x = t
⇒ x = t²
⇒ dx = 2t dt
= ∫ tan^-1 t (2t dt)
= 2 ∫ tan^-1 t t dt
= 2 [tan^-1 t . \(\frac{t^2}{2}\) – ∫ \(\frac{1}{1+t^2} \cdot \frac{t^2}{2}\)] + c
= 2 \(\left[\frac{t^2}{2} \tan ^{-1} t-\frac{1}{2} \int \frac{1+t^2-1}{1+t^2} d t\right]\)
= t² tan^-1 t – ∫ (1 – \(\frac{1}{1+t^2}\)) dt
= t² tan^-1 t – t + tan^-1 t + c
= (1 + t²) tan^-1 t – t + c
= (1 + x) tan^-1 √x – √x + c

(ii) Let I = ∫ x³ tan^-1 (x²) dx
put x² = t
⇒ 2x dx = dt
∴ I = ∫ t . tan^-1 t \(\frac{dt}{2}\)

Question 15.
(i) ∫ sin³ √x dx
(ii) ∫ \(\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}\) dx (NCERT)
Solution:
(i) Let I = ∫ sin³ √x dx
put √x = t
⇒ x = t²
⇒ dx = 2t dt
= ∫ sin³ t (2t dt)
= 2 ∫ \(\frac{t}{4}\) [3 sin t – sin 3t] dt
[∵ sin 3t = 3 sin t – 4 sin³ t]
⇒ sin³ t = \(\frac{1}{4}\) [3 sin t – sin 3t]

(ii) Let I = ∫ \(\frac{x \cos ^{-1} x}{\sqrt{1-x^2}}\) dx
put cos^-1 x = t
x = cos t
dx = – sin t dt
= ∫ \(\frac{\cos t \cdot t(-\sin t d t)}{\sqrt{1-\cos ^2 t}}\)
= – ∫ t cos tdt
= – [t sin t – ∫ 1 . sin t dt] + c
= – [t sin t + cos t] + c
= – [\(\sqrt{1-x^2}\) cos^-1 x + x] + c

Question 16.
(i) ∫ sin^-1 (3x – 4x³) dx
(ii) ∫ tan^-1 \(\left(\frac{3 x-x^3}{1-3 x^2}\right)\) dx
Solution:
(i) Let I = ∫ sin^-1 (3x – 4x³) dx
put x = sin θ
⇒ dx = cos θ dθ
= ∫ sin^-1 (3 sin θ – 4 sin³ θ) cos θ dθ
= ∫ sin^-1 (sin 3θ) cos θ dθ
= 3 ∫ θ . cos θ dθ
= 3 [θ sin θ – ∫ 1 . sin θ] + c
= 3 [θ sin θ + cos θ] + c
= 3 [sin^-1 x . x + \(\sqrt{1-x^2}\)] + c
[∵ cos θ = \(\sqrt{1-\sin ^2 \theta}\)
= \(\sqrt{1-x^2}\)]
= 3 [x sin^-1 x + \(\sqrt{1-x^2}\)] + c

(ii) Let I = ∫ tan^-1 \(\left(\frac{3 x-x^3}{1-3 x^2}\right)\) dx
put x = tan θ
⇒ dx = sec² θ dθ
= ∫ tan^-1 \(\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)\) sec² θ dθ
= ∫ tan^-1 (tan 3θ) sec² θ dθ
= 3 ∫ θ sec² θ dθ

Question 17.
(i) ∫ \(\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}\) dx
(ii) ∫ \(\frac{x^2 \tan ^{-1} x}{1+x^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}\) dx
put x = sin θ
⇒ dx = cos θ dθ
∴ θ = sin^-1 x
⇒ I = ∫ \(\frac{\theta \cos \theta d \theta}{\left(1-\sin ^2 \theta\right)^{3 / 2}}\)
= \(\frac{\theta \cos \theta d \theta}{\cos ^3 \theta}\)
= ∫ θ sec² θ dθ
= θ tan θ – ∫ 1 . tan θ dθ
= θ tan θ – ∫ \(\frac{\sin \theta}{\cos \theta}\) dθ
= θ tan θ + log |cos θ| + C

(ii) Let I = ∫ \(\frac{x^2 \tan ^{-1} x}{1+x^2}\) dx
put tan^-1 x = θ
⇒ x = tan θ
⇒ dx = sec² θ dθ
= ∫ \(\frac{\tan ^2 \theta \cdot \theta}{1+\tan ^2 \theta}\) . sec² θ dθ
= ∫ θ tan² θ dθ
= ∫ θ (sec² θ – 1) dθ

= ∫ θ sec² θ dθ – ∫ θ dθ
= θ tan θ – ∫ 1 . tan θ dθ – \(\frac{\theta^2}{2}\) + c
= θ tan θ + log |cos θ| – \(\frac{\theta^2}{2}\) + c
= x tan^-1 x + log \(\left|\frac{1}{\sqrt{1+x^2}}\right|\) – \(\frac{1}{2}\) (tan^-1 x)² + c
= x tan^-1 x – \(\frac{1}{2}\) log |1 + x²| – \(\frac{1}{2}\) (tan^-1 x)² + c

Question 18.
(i) ∫ e^x sin x dx
(ii) ∫ e^2x sin x dx
Solution:
(i) Let I = ∫ e^x sin x dx
= sin x e ^x – ∫ cos x e^x dx
= e^x sin x – [cos x . e^x – ∫ – sin x e^x dx]
= (sin x – cos x) e^x – 1
⇒ 2I = (sin x – cos x) e^x – 1
⇒ I = \(\frac{e^x}{2}\) (sin x – cos x) + C

[Students can also take first function as sin x]
= – e^2x cos x + ∫ 2 e^2x cos x dx
∴ I = – e^2x cos x + 2 [e^2x sin x – ∫ 2 e^2x sin x dx]
⇒ I = e^2x (- cos x + 2 sin x) – 4 I
⇒ I = \(\frac{e^{2 x}}{5}\) [2 sin x – cos x] + C

Question 19.
(i) ∫ e^ax sin bx dx
(ii) ∫ e^ax cos (bx + c) dx
Solution:
(i) Let I = ∫ e^ax sin bx dx

(ii) Let I = ∫ e^ax cos (bx + c) dx
= cos (bx + c) \(\frac{e^{a x}}{a}\) + ∫ sin (bx + c) . b . \(\frac{e^{a x}}{a}\) dx

Question 20.
(i) ∫ x² e^x3 cos x³ dx
(ii) ∫ e^x sin² x dx
Solution:
(i) Let I = ∫ x² e^x3 cos x³ dx
put x³ = t
⇒ 3x² dx = dt
= ∫ e^t cos t \(\frac{d t}{3}\)
= \(\frac{1}{3}\) ∫ e^t cos t dt
= \(\frac{1}{3}\) I₁ ………………(1)
where I₁ = ∫ e^t cos t dt
= e^t sin t – ∫ e^t sin t dt
= e^t sin t – [- e^t cos t – ∫ e^t (- cos t) dt]
∴ I₁ = e^t sin t + e^t cos t – I₁
⇒ I₁ = \(\frac{e^t}{2}\) [sin t + cos t]
∴ from (1) ;
∴ I = \(\frac{e^t}{6}\) [sin t + cos t]
I = \(\frac{e^{x^3}}{6}\) [sin x³ + cos x³] + c

(ii) Let I₁ = ∫ e^x sin² x dx

Question 21.
(i) ∫ e^{sin-1 x} dx
(ii) ∫ \(\frac{e^{m \tan ^{-1} x}}{\left(1+x^2\right)^{3 / 2}}\) dx
(iii) ∫ cosec³ x dx
Solution:
(i) put sin^-1 x = t
⇒ x = sin t
⇒ dx = cos t dt
∴ I = ∫ e^{sin-1 x} dx
= ∫ e^t cos t dt
= e^t sin t – ∫ e^t sin t dt
∴ I = e^t sin t – [e^t (- cos t) + ∫ e^t cos t dt]
⇒ I = e^t (sin t + cos t) – I
⇒ I = \(\frac{e^t}{2}\) (sin t + cos t) + C
⇒ I = \(\frac{e^{\sin ^{-1} x}}{2}\left[x+\sqrt{1-x^2}\right]\) + C

(ii) Let I = ∫ \(\frac{e^m \tan ^{-1} x}{\left(1+x^2\right)^{3 / 2}}\) dx
put tan^-1 x = t
⇒ x = tan t
⇒ dx = sec² t dt

(iii) Let I = ∫ cosec³ x dx
= ∫ cosec x cosec² x dx
= cosec x (- cot x) – ∫ – cot x cosec x (- cot x) dx
= – cot x cosec x – ∫ cosec x (cosec² x – 1) dx
= – cot x cosec x – ∫ cosec³ x dx + ∫ cosec x dx
⇒ I = – cot x cosec x – I + ∫ cosec x dx
⇒ 2I = – cot x cosec x + log |tan \(\frac{x}{2}\)| + c
⇒ I = \(\frac{-\cot x \ {cosec} x}{2}+\frac{1}{2} \log \left|\tan \frac{x}{2}\right|\) + c

Question 22.
(i) ∫ cos (log x) dx
(ii) ∫ \(\frac{\sin ^{-1} x}{x^2}\) dx
Solution:
(i) Let I = ∫ cos (log x) . 1 dx
= cos (log x) . x – ∫ – sin (log x) . \(\frac{1}{x}\) . x dx
= x cos (log x) + ∫ sin (log x) . 1 dx
= x cos (log x) + sin (log x) . x – ∫ cos (log x) . \(\frac{1}{x}\) . x dx
∴ I = x [cos (log x) + sin (log x)] – I
⇒ 2I = x [cos (log x) + sin (log x)]
⇒ I = \(\frac{x}{2}\) [cos (log x) + sin (log x)] + c

(ii) Let I = ∫ \(\frac{\sin ^{-1} x}{x^2}\) dx
put sin^-1 x = t
x = sin t
⇒ dx = cos t dt
= ∫ \(\frac{t}{\sin ^2 t}\) cos t dt
= ∫ t (cot t cosec t dt)
= t (- cosec t) – ∫ 1 . (- cosec t) dt + c

= – t cosec t + ∫ cosec t dt + c
= – t cosec t – log |cosec t + cot t| + c
= \(-\frac{1}{x} \sin ^{-1} x-\log \left|\frac{1}{x}+\frac{\sqrt{1-x^2}}{x}\right|\) + c
= \(\frac{-\sin ^{-1} x}{x}-\log \left|\frac{1+\sqrt{1-x^2}}{x}\right|\) + c

Accessing ISC Maths Class 12 Solutions Chapter 8 Integrals Ex 8.10 can be a valuable tool for students seeking extra practice.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.10

Question 1.
(i) ∫ \(\frac{x-1}{(x-2)(x-3)}\) dx
(ii) ∫ \(\frac{3 x+5}{x^2+3 x-18}\) dx
Solution:
(i) Let \(\frac{x-1}{(x-2)(x-3)}\) = \(\frac{\mathrm{A}}{x-2}\) + \(\frac{\mathrm{B}}{x-3}\) ……….(1)
Multiplying both sides of eqn. (1) by (x – 2) (x – 3) ; we have
x – 1 = A (x – 3) + B (x – 2) ……………(2)
putting x = 2, 3 successively in eqn. (2) ; we have
1 = – A
A = – 1
and 2 = B
∴ from (1); we have
\(\int \frac{x-1}{(x-2)(x-3)} d x=\int \frac{-1}{x-2} d x+\int \frac{2 d x}{x-3}\)
= – log |x – 2| + 2 log |x – 3| + C

(ii) Let \(\frac{3 x+5}{x^2+3 x-18}=\frac{3 x+5}{(x-3)(x+6)}\)
= \(\frac{\mathrm{A}}{x-3}+\frac{\mathrm{B}}{x+6}\) …………..(1)
Multiplying both sides of eqn. (1) by (x – 3) (x + 6) ; we have
3x + 5 = A (x + 6) + B (x – 3) …………….(2)
putting x = 3, – 6 successively in eqn. (2) ;
14 = 9A
⇒ A = \(\frac{14}{9}\)
and – 13 = – 9B
⇒ B = \(\frac{13}{9}\)
∴ from (1) ;
\(\int \frac{3 x+5}{x^2+3 x-18} d x=\frac{14}{9} \int \frac{d x}{x-3}+\frac{13}{9} \int \frac{d x}{x+6}\)
= \(\frac{14}{9}\) log |x – 3| + \(\frac{13}{9}\) log |x – 6| + C

Question 1 (old).
(i) ∫ \(\frac{d x}{(x+1)(x+2)}\) (NCERT)
(ii) ∫ \(\frac{x}{(x+1)(x+2)}\) (NCERT)
Solution:
(i) Let \(\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\) ………….(1)
Multiply both sides of eqn. (1) by (x + 1) (x + 2) ; we get
I = A (x + 2) + B (x + 1) ……………….(2)
putting x = – 1, – 2 successively in eqn. (2) ; we have
1 = A and 1 = – B
⇒ B = – 1
∴ from (1) ; we have
\(\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}-\frac{1}{x+2}\)
I = \(\frac{d x}{(x+1)(x+2)}\)
= ∫ \(\frac{d x}{x+1}\) – ∫ \(\frac{d x}{x+2}\)
= log |x + 1| – log |x + 2| + C
= log \(\left|\frac{x+1}{x+2}\right|\) + C

(ii) Let \(\frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\) …………..(1)
Multiplying eqn. (1) by (x + 1) (x + 2) ; we have
x = A (x + 2) + B (x + 1) ……………(2)
putting x = – 1 in eqn. (2) ;
– 1 = A
putting x = – 2 in eqn. (2) ; we have
– 2 = – B
⇒ B = 2
∴ eqn. (1) gives ;
∴ \(\int \frac{x}{(x+1)(x+2)} d x=-\int \frac{d x}{x+1}+2 \int \frac{d x}{x+2}\)
= – log |x + 1| + 2 log |x + 2| + C
= log \(\frac{(x+2)^2}{|x+1|}\) + C

Question 2.
(i) ∫ \(\frac{2 x+7}{x^2-x-2}\) dx (ISC 2020)
(ii) ∫ \(\frac{x^2+1}{x^2-5 x+6}\) dx
Solution:
(i) Let \(\frac{2 x+7}{x^2-x-2}\) = \(\frac{2 x+7}{(x-2)(x+1)}\)
= \(\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x+1}\) …………….(1)
Multiply both sides of eqn. (1) by (x – 20 (x + 1) ; we have
2x + 7 = A (x + 1) + B (x – 2) ………….(2)
putting x = – 1 in eqn. (2) ; we have
11 = 3A
A = \(\frac{11}{3}\)
∴ from (1) ; we have
∫ \(\frac{2 x+7}{x^2-x-2}\) dx = \(\int \frac{11 / 3}{x-2} d x+\int \frac{-5 / 3}{x+1} d x\)
= \(\frac{11}{3}\) log |x – 2| – \(\frac{5}{3}\) log |x – 1| + C

(ii) Let I = ∫ \(\frac{x^2+1}{x^2-5 x+6}\) dx
= ∫ \(\left[1+\frac{5 x-5}{x^2-5 x+6}\right]\) dx
= x + ∫ \(\frac{5(x-1) d x}{(x-2)(x-3)}\)
= x + 5I₁
where I₁ = ∫ \(\frac{(x-1) d x}{(x-2)(x-3)}\)
Let \(\frac{x-1}{(x-2)(x-3)}=\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-3}\) ……………(1)
Multiplying both sides of eqn. (1) by (x – 2) (x – 3) ; we have
x – 1 = A (x – 3) + B (x – 2) …………..(2)
putting x = 2, 3 successively in eqn. (2)
1 = – A
⇒ A = – 1
and 2 = B
∴ from (1) ;
I₁ = ∫ \(\frac{-1}{x-2}\) dx + ∫ \(\frac{2}{x-3}\) dx
= – log |x – 2| + 2 log |x – 3|
Thus I = x – 5 log |x – 2| + 10 log |x- 3| + C

Question 2 (old).
(i) ∫ \(\frac{x}{(x-1)(x-2)}\) dx (NCERT)
(ii) ∫ \(\frac{x^2+1}{x^2-5 x+6}\) dx (NCERT)
Solution:
(i) Let \(\frac{x}{(x-1)(x-2)}\) = \(\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}\) ……….(1)
Multiplying both sides eqn. (1) by (x – 1) (x – 2) ; we get
x = A (x – 2) + B (x – 1) ………….(2)
putting x = 1, 2 successively in eqn. (2) ; we have
1 = A (- 1)
⇒ A = – 1
and 2 = B
∴ from (1) ; we get
\(\frac{x}{(x-1)(x-2)}=-\frac{1}{x-1}+\frac{2}{x-2}\)
∴ ∫ \(\frac{x d x}{(x-1)(x-2)}\) = – ∫ \(\frac{1}{x-1}\) dx + 2 ∫ \(\frac{1}{x-2}\) dx
= – log |x – 1| + 2 log |x – 2| + C

(ii) Let \(\frac{x^2+1}{(x-2)(x-3)}\) = 1 + \(\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-3}\) ……….(1)
Multiplying both sides eqn. (1) by (x – 2) (x – 3) ; we get
x² + 1 = (x – 2) (x – 3) + A (x – 3) + B (x – 2) ………..(2)
putting x = 2 in eqn. (2) ;
– 5 = A
putting x = 3 in eqn. (2)
10 = B
∴ from (1) ; we have
\(\frac{x^2+1}{(x-2)(x-3)}\) = 1 – \(\frac{5}{x-2}+\frac{10}{x-3}\)
∴ ∫ \(\frac{\left(x^2+1\right) d x}{(x-2)(x-3)}\) = x – 5 log |x – 2| + 10 log |x – 3| + C

Question 3.
(i) ∫ \(\frac{x^3+x+1}{x^2-1}\) dx (NCERT)
(ii) ∫ \(\frac{x^3+1}{x^3-x}\) dx
Solution:
(i) Let \(\frac{x^3+x+1}{x^2-1}\) = x + \(\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x+1}\) ………….(1)
Multiply eqn. (1) by (x² – 1) ; we get
x³ + x + 1 = x (x² – 1) + A (x + 1) B (x – 1) …………..(2)
putting x = 1 in eqn. (2) ;
3 = 2A
⇒ A = \(\frac{3}{2}\)
putting x = – 1 in eqn. (2) ;
– 1 = – 2B
⇒ B = \(\frac{1}{2}\)
∴ from (1) ; we have
\(\frac{x^3+x+1}{x^2-1}=x+\frac{\frac{3}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}\)
∴ ∫ \(\frac{x^3+x+1}{x^2+1}\) dx = \(\frac{x^2}{2}\) + \(\frac{3}{2}\) log |x – 1| + \(\frac{1}{2}\) log |x + 1| + C

(ii) Let I = ∫ \(\frac{x^3+1}{x^3-x}\) dx
= ∫ \(\frac{x^3-x+x+1}{x^3-x}\) dx
= ∫ \(\left[1+\frac{x+1}{x\left(x^2-1\right)}\right]\) dx
= ∫ \(\left[1+\frac{x+1}{x(x-1)(x+1)}\right]\) dx
= ∫ \(\left[1+\frac{1}{x(x-1)}\right]\) dx
= ∫ \(\left[1+\frac{-1}{x}+\frac{1}{x-1}\right]\) dx
= x – log |x| + log |x – 1| + C

Question 3 (old).
(ii) ∫ \(\frac{1-x^2}{x(1-2 x)}\) dx (NCERT)
Solution:
Let \(\frac{1-x^2}{x(1-2 x)}\) = \(\frac{1}{2}+\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{1-2 x}\) ………….(1)
Multiply eqn. (1) by x (1 – 2x) ;
1 – x² = \(\frac{1}{2}\) x (1 – 2x) + A (1 – 2x) + B
putting x = 0 in eqn. (2) ;
1 = A
putting x = \(\frac{1}{2}\) in eqn. (2) ;
\(\frac{3}{4}=\frac{B}{2}\)
B = \(\frac{3}{2}\)
∴ eqn. (1) gives ;
\(\frac{1-x^2}{x(1-2 x)}=\frac{1}{2}+\frac{1}{x}+\frac{\frac{3}{2}}{1-2 x}\)
∴ ∫ \(\frac{\left(1-x^2\right)}{x(1-2 x)}\) dx = \(\frac{1}{2}\) x + log |x| + \(\frac{3}{2} \log \frac{|1-2 x|}{-2}\) + C
= \(\frac{x}{2}\) + log |x| – \(\frac{3}{4}\) log |1 – 2x| + C

Question 4.
(i) ∫ \(\frac{x}{(x-1)(x-2)(x-3)}\) dx (NCERT)
(ii) ∫ \(\frac{3 x-1}{(x-1)(x-2)(x-3)}\) dx (NCERT)
Solution:
(i) Let \(\frac{x}{(x-1)(x-2)(x-3)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}+\frac{\mathrm{C}}{x-3}\) ………(1)
Multiplying eqn. (1) by (x – 1) (x – 2) (x – 3) ; we get
x = A (x – 2) (x – 3) + B (x – 1) (x – 3) + C (x – 1) (x – 2) …………(2)
putting x = 1 in eqn. (2) ;
1 = 2 A
⇒ A = \(\frac{1}{2}\)
putting x = 1 in eqn. (2) ;
2 = – B
⇒ B = – 2
putting x = 3 in eqn. (2) ;
3 = 2 C
⇒ C = \(\frac{3}{2}\)
∴ eqn. (1) becomes,
\(\frac{x}{(x-1)(x-2)(x-3)}=\frac{\frac{1}{2}}{x-1}-\frac{2}{x-2}+\frac{\frac{3}{2}}{x-3}\)
∴ \(\int \frac{x d x}{(x-1)(x-2)(x-3)}=\frac{1}{2} \int \frac{d x}{x-1}-2 \int \frac{d x}{x-2}+\frac{3}{2} \int \frac{d x}{x-3}\)
= \(\frac{1}{2}\) log |x – | – 2 log |x – 2| + \(\frac{3}{2}\) log |x – 3| + C

(ii) Let \(\frac{3 x-1}{(x-1)(x-2)(x-3)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}+\frac{\mathrm{C}}{x-3}\) ……….(1)
Multiplying eqn. (1) by (x – 1) (x – 2) (x – 3) ; we have
3x – 1 = A (x – 2) (x – 3) + B (x – 1) (x – 3) + C (x – 1) (x – 2) …………..(2)
putting x = 1 in eqn. (2) ;
2 = 2A
⇒ A = 1 ;
putting x = 2 in eqn. (2) ;
5 = – B
⇒ B = – 5
putting x = 3 in eqn. (2) ;
8 = 2C
⇒ C = 4
∴ eqn. (1) gives ;
\(\int \frac{(3 x-1) d x}{(x-1)(x-2)(x-3)}=\int \frac{d x}{x-1}-\int \frac{5 d x}{x-2}+4 \int \frac{d x}{x-3}\)
= log |x – 1| – 5 log |x – 2| + 4 log |x – 3| + C

Question 5.
(i) ∫ \(\frac{5 x}{(x+1)\left(x^2-4\right)}\) dx
(ii) ∫ \(\frac{x}{(x-1)^2(x+2)}\) dx (NCERT)
Solution:
(i) Let \(\frac{5 x}{(x+1)\left(x^2-4\right)}\) = \(\frac{5 x}{(x+1)(x-2)(x+2)}\)
i.e. \(\frac{5 x}{(x+1)(x-2)(x+2)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x-2}+\frac{\mathrm{C}}{x+2}\) ……………..(1)
Multiplying eqn. (1) by (x + 1) (x – 2) (x + 2) ; we get
5x = A (x² – 4) + B (x + 1) (x + 2) + C (x + 1) (x – 2) ………..(2)
putting x = – 1 in eqn. (2) ;
– 5 = – 3A
⇒ A = \(\frac{5}{3}\)
putting x = 2 in eqn. (2) ;
10 = 12 B
⇒ B = \(\frac{5}{6}\)
putting x = – 2 in eqn. (2) ;
– 10 = 4C
⇒ C = – \(\frac{5}{2}\)
∴ eqn. (1) gives ;
\(\frac{5 x}{(x+1)\left(x^2-4\right)}=\frac{\frac{5}{3}}{x+1}+\frac{\frac{5}{6}}{x-2}-\frac{\frac{5}{2}}{x+2}\)
∴ ∫ \(\frac{5 x d x}{(x+1)\left(x^2-4\right)}\) = \(\frac{5}{3} \int \frac{d x}{x+1}+\frac{5}{6} \int \frac{d x}{x-2}-\frac{5}{2} \int \frac{d x}{x+2}\)
= \(\frac{5}{3}\) log |x + 1| + \(\frac{5}{6}\) log |x – 2| – \(\frac{5}{2}\) log |x + 2| + C

(ii) Let \(\frac{x}{(x-1)^2(x+2)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{(x-1)^2}+\frac{\mathrm{C}}{x+2}\) ………….(1)
Then x = A (x – 1) (x + 2) + B (x + 2) + C (x – 1)² …………..(2)
putting x = 1, – 2, and 0 successively in eqn. (2) ; we get
1 = 3B
⇒ B = 1/3
– 2 = 9C
⇒ C = – 2/9
and 0 = – 2A + 2B + 2C
⇒ 2A = \(\frac{2}{3}-\frac{2}{9}\)
A = 2/9
From (1) ; we have

Question 6.
(i) ∫ \(\frac{3 x-1}{(x-2)^2}\) dx
(ii) ∫ \(\frac{3 x+5}{x^3-x^2-x+1}\) dx (NCERT)
Solution:
(i) Let \(\frac{3 x-1}{(x-2)^2}\) = \(\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{(x-2)^2}\) ………..(1)
Multiply both sides of eqn. (1) by (x – 2)² ; we have
3 = A ;
– 1 = – 2A + B
⇒ B = 2
∴ from (1) ;
\(\int \frac{3 x-1}{(x-2)^2} d x=\int \frac{3}{x-2} d x+\int \frac{5}{(x-2)^2} d x\)
= 3 log |x – 2| + 5 \(\frac{(x-2)^{-2+1}}{(-2+1)}\) + C
= 3 log |x – 2| – \(\frac{5}{x-2}\) + C

(ii) \(\frac{3 x+5}{x^3-x^2-x+1}=\frac{3 x+5}{(x-1)\left(x^2-1\right)}=\frac{3 x+5}{(x-1)^2(x+1)}\)
Let \(\frac{3 x+5}{(x-1)^2(x+1)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x-1}+\frac{\mathrm{C}}{(x-1)^2}\) ……………..(1)
Multiplying eqn. (1) by (x – 1)² (x + 1) ; we get
3x + 5 = A (x – 1)² + B (x² – 1) + C (x + 1) ……………(2)
putting x = 1 in eqn. (2) ;
8 = 2C
⇒ C = 4
putting x = – 1 in eqn. (2) ;
2 = 4A
⇒ A = \(\frac{1}{2}\)
Coeff. of x² ;
0 = A + B
⇒ B = – \(\frac{1}{2}\)
∴ eqn. (1) gives ;
\(\int \frac{(3 x+5) d x}{x^3-x^2-x+1}=\frac{1}{2} \int \frac{d x}{x+1}-\frac{1}{2} \int \frac{d x}{x-1}+4 \int \frac{d x}{(x-1)^2}\)
= \(\frac{1}{2}\) log |x + 1| – \(\frac{1}{2}\) log |x – 1| – \(\frac{4}{x-1}\) + C
= \(\frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}\) + C, x ≠ 1.

Question 7.
(i) ∫ \(\frac{2}{(1-x)\left(1+x^2\right)}\) dx
(ii) ∫ \(\frac{5 x}{(x+1)\left(x^2+9\right)}\) dx (NCERT)
(iii) ∫ \(\frac{4}{(x-2)\left(x^2+4\right)}\) dx
Solution:
(i) Let \(\frac{2}{(1-x)\left(1+x^2\right)}=\frac{\mathrm{A}}{1-x}+\frac{\mathrm{B} x+\mathrm{C}}{1+x^2}\) …………(1)
Multiplying eqn. (1) by (1 – x) (1 + x²) ; we get
2 = A (1 + x²) + (Bx + C) (1 – x)
put x = 1 in eqn. (2) ;
2 = 2A
⇒ A = 1
Coeff. of x² ;
0 = A – B
⇒ B = 1
Coeff. of x ;
0 = B – C
⇒ C = 1
∴ From (1) ; we get
\(\int \frac{2 d x}{(1-x)\left(1+x^2\right)}=\int \frac{d x}{1-x}+\int \frac{(x+1)}{x^2+1} d x\)
= \(\frac{\log |1-x|}{-1}+\frac{1}{2} \int \frac{2 x d x}{x^2+1}+\int \frac{d x}{x^2+1}\)
= – log |1 – x| + \(\frac{1}{2}\) log |x² + 1| + tan^-1 x + C

(ii) Let \(\frac{5 x}{(x+1)\left(x^2+9\right)}=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+9}\) …………(1)
Multiplying eqn. (1) by (x + 1) (x² + 9) ; we get
5x = A (x² + 9) + (Bx + C) (x + 1)
putting x = – 1 in eqn. (2) ; we get
– 5 = 10A
⇒ A = – \(\frac{1}{2}\)
Coeff. of x² ;
0 = A + B
⇒ B = \(\frac{1}{2}\)
Coeff. of x ;
5 = B + C
⇒ C = \(\frac{9}{2}\)
∴ From eqn. (1) ; we have

(iii) Let I = ∫ \(\frac{4}{(x-2)\left(x^2+4\right)}\) dx
Let \(\frac{4}{(x-2)\left(x^2+4\right)}=\frac{A}{x-2}+\frac{B x+C}{x^2+4}\) …………(1)
Multiply both sides of eqn. (1) by (x – 2) (x² + 4) ; we have
4 = A (x² + 4) + (Bx + C) (x – 2) …………….(2)
put x = 2 in eqn. (2) ; we have
4 = 8A
⇒ A = \(\frac{1}{2}\)
Coeff. of x² ;
0 = A + B
⇒ B = \(\frac{1}{2}\)
Coeff. of x ;
0 = – 2B + C
⇒ C = 2B
= 2 (- \(\frac{1}{2}\)) = – 1
∴ From eqn. (1) ; we have
I = \(\int \frac{\frac{1}{2}}{x-2} d x+\int \frac{-\frac{x}{2}-1}{x^2+4} d x\)
= \(\frac{1}{2}\) log |x – 2| – \(\frac{1}{2} \int \frac{x d x}{x^2+4}-\int \frac{d x}{x^2+4}\) + C
= \(\frac{1}{2} \log |x-2|-\frac{1}{4} \log \left|x^2+4\right|-\frac{1}{2} \tan ^{-1} \frac{x}{2}\) + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C
and ∫ \(\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]

Question 7 (old).
(i) ∫ \(\frac{x}{(x-1)\left(x^2+1\right)}\) dx (NCERT)
(ii) ∫ \(\frac{x^2+x+1}{(x+2)\left(x^2+1\right)}\) dx (NCERT)
Solution:
(i) Let \(\frac{x}{(x-1)\left(x^2+1\right)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+1}\) …………(1)
Multiply both sides of eqn. (1) by (x – 1) (x² + 1) ; we have
x = A (x² + 1) + (Bx + C) (x – 1)
putting x = 1 in eqn. (2) ; we have
1 = 2A
⇒ A = \(\frac{1}{2}\)
Coefficients of x² ;
0 = A + B
⇒ B = \(-\frac{1}{2}\)
Coefficients of x ;
1 = – B + C
⇒ C = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∴ From eqn. (1) ; we have

(ii) Let \(\frac{x^2+x+1}{(x+2)\left(x^2+1\right)}=\frac{\mathrm{A}}{x+2}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+1}\) ………..(1)
Multiplying both sides of eqn. (1) by (x² + 1) (x + 2) ; we get
x² + x + 1 = A (x² + 1) + (Bx + C) (x + 2)
putting x = – 2 in eqn. (2) ; we have
4 – 2 + 1 = A (4 + 1)
⇒ A = \(\frac{3}{5}\)
Coefficients of x² ;
1 = A + B
⇒ B = 1 – \(\frac{3}{5}\) = \(\frac{2}{5}\)
Coefficients of x ;
1 = 2B + C
⇒ C = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
∴ From eqn. (1) ; we have

Question 8.
(i) ∫ \(\frac{x^2+x}{x^3-x^2+x-1}\) dx
(ii) ∫ \(\frac{x^3}{(x-1)\left(x^2+1\right)}\) dx
Solution:
(i) Let \(\frac{x^2+x}{x^3-x^2+x-1}=\frac{x^2+x}{\left(x^2+1\right)(x-1)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+1}\) ……………….(1)
Multiply both sides of eqn. (1) by (x² + 1) (x – 1) ; we have
x² + x = A (x² + 1) + (Bx + C) (x – 1) …………….(2)
putting x = 1 in eqn. (2) ; we have
1 + 1 = A (1 + 1)
⇒ A = 1
Coefficients of x² ;
1 = A + B
⇒ B = 0
Coefficients of x ;
1 = – B + C
⇒ C = 1
∴ from (1) ; we get
\(\frac{x^2+x}{x^3-x^2+x-1}=\frac{1}{x-1}+\frac{1}{x^2+1}\)
∴ \(\int \frac{\left(x^2+x\right) d x}{x^3-x^2+x-1}=\int \frac{1}{x-1} d x+\int \frac{d x}{x^2+1}\)
= log |x – 1| + tan^-1 x + C
[∵ ∫ \(\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]

(ii) Let I = ∫ \(\frac{x^3}{(x-1)\left(x^2+1\right)}\)
Here the integrand is not a proper function.
So on dividing x³ by (x – 1) (x² + 1) ; we get
quotient = 1
and remainder = x² – x + 1
∴ \(\frac{x^3}{(x-1)\left(x^2+1\right)}=1+\frac{x^2-x+1}{(x-1)\left(x^2+1\right)}\) …………..(1)
Let \(\frac{x^2-x+1}{(x-1)\left(x^2+1\right)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+1}\) …………….(2)
Multiplying both sides of eqn. (1) by (x – 1) (x² + 1) ; we get
x² – x + 1 = A (x² + 1) + (Bx + C) (x – 1) ……………(3)
putting x = 1 in eqn. (3) ; we have
1 – 1 + 1 = A (2)
⇒ A = \(\frac{1}{2}\)
Coefficients of x² ;
1 = A + B
⇒ B = \(\frac{1}{2}\)
Coefficients of x ;
– 1 = – B + C
⇒ C = – 1 + \(\frac{1}{2}\) = – \(\frac{1}{2}\)

Question 9.
(i) ∫ \(\frac{d x}{1-x^3}\)
(ii) ∫ \(\frac{d x}{(x+1)^2\left(x^2+1\right)}\)
Solution:
(i) \(\frac{1}{1-x^3}=\frac{1}{(1-x)\left(1+x+x^2\right)}\)
Let \(\frac{1}{(1-x)\left(1+x+x^2\right)}=\frac{\mathrm{A}}{1-x}+\frac{\mathrm{B} x+\mathrm{C}}{x^2+x+1}\) ………….(1)
Multiplying eqn. (1) by (1 – x) (x² + x + 1) ; we get
1 = A (x² + x + 1) + (Bx + C) (1 – x) ………….(2)
putting x = 1 in eqn. (2);
1 = 3A
⇒ A = \(\frac{1}{3}\)
Coefficients of x² ;
0 = A – B
⇒ B = \(\frac{1}{3}\)
Coefficients of x ;
0 = A + B – C
⇒ C = \(\frac{2}{3}\)
From (1) ; we have

Question 10.
(i) ∫ \(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\) dx
(ii) ∫ \(\frac{x^2}{\left(1+x^3\right)\left(2+x^3\right)}\) dx
(iii) ∫ \(\frac{2 x}{\left(x^2+1\right)\left(x^2+2\right)^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)}\) dx
put x² = t
⇒ 2x dx = dt
∴ I = ∫ \(\frac{d t}{(t+1)(t+3)}\)
Let \(\frac{1}{(t+1)(t+3)}=\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+3}\) …………….(1)
⇒ 1 = A (t + 3) + B (t + 1) ………….(2)
putting t = – 1, – 3 successively in eqn. (2) ; we get
1 = 2A
⇒ A = 1/2
and 1 = – 2B
⇒ B = – 1/2
∴ from (1) ;
\(\frac{1}{(t+1)(t+3)}=\frac{1 / 2}{t+1}+\frac{-1 / 2}{t+3}\)
∴ I = \(\int \frac{\frac{1}{2}}{t+1} d t-\frac{1}{2} \int \frac{d t}{t+3}\)
= \(\frac{1}{2}\) log |t + 1| – \(\frac{1}{2}\) log |t + 3| + c
= \(\frac{1}{2} \log \left|\frac{t+1}{t+3}\right|\) + c
= \(\frac{1}{2} \log \left|\frac{x^2+1}{x^2+3}\right|\) + c

(ii) Let I = ∫ \(\frac{x^2 d x}{\left(1+x^3\right)\left(2+x^3\right)}\) ;
put x³ = y
⇒ 3x² dx = dy
= \(\frac{1}{3} \int \frac{d y}{(y+1)(y+2)}\) …………..(1)
Let \(\frac{1}{(y+1)(y+2)}=\frac{\mathrm{A}}{y+1}+\frac{\mathrm{B}}{y+2}\) ………….(2)
Multiplying eqn. (2) by (y + 1) (y + 2) ; we get
1 = A (y + 2) + B (y + 1) ……………..(3)
putting y = – 1 in eqn. (3) ;
1 = A
putting y = – 2in eqn. (3) ;
1 = – B
⇒ B = – 1
∴ From (2) ; we have
\(\frac{1}{(y+1)(y+2)}=\frac{1}{y+1}-\frac{1}{y+2}\)
∴ \(\int \frac{d y}{(y+1)(y+2)}=\int \frac{d y}{y+1}-\int \frac{d y}{y+2}\)
∴ From (1) ; we have
1 = \(\frac{1}{3}\) [log |y + 1| – kog |y + 2|] + C
= \(\frac{1}{3} \log \left|\frac{x^3+1}{x^3+2}\right|\) + C

(iii) Let I = ∫ \(\frac{2 x d x}{\left(x^2+1\right)\left(x^2+2\right)^2}\) dx
put x² = t
⇒ 2x dx = dt
= ∫ \(\frac{d t}{(t+1)(t+2)^2}\)
Let \(\frac{1}{(t+1)(t+2)^2}\) = \(\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+2}+\frac{\mathrm{C}}{(t+2)^2}\) …………….(1)
Multiply both sides of eqn. (1) by (t + 1) (t + 2)² ; we have
1 = A (t + 2)² + B (t + 1) (t + 2) + C (t + 1) ………….(2)
putting t = – 1 in eqn. (2) ; we have
1 = A (1)²
⇒ A = 1
putting t = – 2 in eqn. (2) ; we have
1 = C (- 2 + 1)
⇒ C = – 1
Coeff. of t² ;
0 = A + B
⇒ B = – A = – 1
∴ from (1) ; we have
∴ I = ∫ \(\frac{1}{(t+1)(t+2)^2}\) dt
= \(\int \frac{d t}{t+1}-\int \frac{d t}{t+2}-\int \frac{d t}{(t+2)^2}\)
= log |t + 1| – log |t + 2| – \(\frac{(t+2)^{-2+1}}{-2+1}\) + C
= log (x² + 1) – log (x² + 2) + \(\frac{1}{x^2+2}\) + C

Question 11.
(i) ∫ \(\frac{d x}{x\left(x^4-1\right)}\) (NCERT)
(ii) ∫ \(\frac{d x}{x\left(x^2+1\right)}\) (NCERT)
Solution:
(i) Let I = ∫ \(\frac{d x}{x\left(x^4-1\right)}\)
put x⁴ = t
⇒ 4x³ dx = dt
⇒ dx = \(\frac{d t}{4 x^3}\)
∴ I = ∫ \(\frac{d t}{4 x^4\left(x^4-1\right)}\)
= \(\frac{1}{4} \int \frac{d t}{t(t-1)}\) …………….(1)
Let \(\frac{1}{t(t-1)}=\frac{\mathrm{A}}{t}+\frac{\mathrm{B}}{t-1}\) ……………(2)
⇒ 1 = A (t – 1) + Bt …………..(3)
putting t = 0, 1 successively in eqn. (3) ; we have
1 = – A
⇒ A = – 1
and 1 = B
⇒ B = 1
∴ From (2) ;
\(\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}\)
∴ from eqn. (1) ; we have
1 = \(\frac{-1}{4}\) log |t| + \(\frac{1}{4}\) log |t – 1| + c
= \(\frac{1}{4} \log \left|\frac{t-1}{t}\right|\) + c
= \(\frac{1}{4} \log \left|\frac{x^4-1}{x^4}\right|\) + c

(ii) Let I = ∫ \(\frac{d x}{x\left(x^2+1\right)}\)
= ∫ \(\)
put x² dx = t
⇒ 2x dx = dt
∴ I = ∫ \(\frac{d t}{2 t(t+1)}\)
Let \(\frac{1}{t(t+1)}=\frac{\mathrm{A}}{t}+\frac{\mathrm{B}}{t+1}\) …………..(1)
Multiplying both sides of eqn. (1) by t (t + 1) ; we have
1 = A (t + 1) + Bt ……………..(2)
putting t = 0, – 1 successively in eqn. (2) ; we have
1 = A
and 1 = – B
B = – 1
∴ from (1) ;

Question 12.
(i) ∫ \(\frac{d x}{x\left(x^n+1\right)}\) (NCERT)
(ii) ∫ \(\frac{d x}{x\left(x^3+1\right)}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{x\left(x^n+1\right)}\)
putting xⁿ = t
n x^n-1 dx = dt
⇒ dx = \(\frac{d t}{n x^{n-1}}\)
∴ I = ∫ \(\frac{d t}{n x^n\left(x^n+1\right)}\)
= \(\frac{1}{n} \int \frac{d t}{t(t+1)}\)
Let \(\frac{1}{t(t+1)}=\frac{\mathrm{A}}{t}+\frac{\mathrm{B}}{t+1}\) ………….(1)
⇒ 1 = A (t + 1) + Bt ………..(2)
putting t = 0, – 1 successively in eqn. (2) ; we have
∴ From (1) ;
\(\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}\)
Thus I = \(\frac{1}{n}\left[\int \frac{1}{t} d t-\int \frac{1}{t+1} d t\right]\)
= \(\frac{1}{n}\) [log |t| – log |t + 1| + c]
= \(\frac{1}{n} \log \left|\frac{t}{t+1}\right|\) + c
= \(\frac{1}{n} \log \left|\frac{x^n}{x^n+1}\right|\) + c

(ii) Let I = ∫ \(\frac{d x}{x\left(x^3+8\right)}\)
putting x³ = t
⇒ 3x² dx = dt
⇒ dx = \(\frac{d t}{3 x^2}\)
∴ I = ∫ \(\frac{d t}{3 x^3\left(x^3+8\right)}\)
= \(\frac{1}{3} \int \frac{d t}{t(t+8)}\)
Let \(\frac{1}{t(t+8)}=\frac{\mathrm{A}}{t}+\frac{\mathrm{B}}{t+8}\) …………….(1)
⇒ 1 = A (t + 8) + Bt …………..(2)
putting t = 0, – 8 successively in eqn. (2) ; we have
1 = 8A
⇒ A = 1/8
and 1 = – 8B
⇒ B = – 1/8
∴ From (1) ;

Question 13.
(i) ∫ \(\frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}\) dx (NCERT)
(ii) ∫ \(\frac{d x}{e^{2 x}+1}\)
Solution:
(i) put e^x = t
⇒ e^x dx = dt
∴ I = ∫ \(\frac{d t}{(1+t)(2+t)}\)
Let \(\frac{1}{(1+t)(2+t)}=\frac{\mathrm{A}}{1+t}+\frac{\mathrm{B}}{2+t}\) ………….(1)
Multiplying eqn. (1) by (1 + t) (2 + t) ; we have
1 = A (2 + t) + B (1 + t) ………..(2)
put t = – 1 in eqn. (2) ;
⇒ 1 = A
put t = – 2 in eqn. (2) ;
1 = – B
⇒ B = – 1
∴ I = \(\int \frac{d t}{1+t}-\int \frac{d t}{2+t}\)
= log |1 + t| – log |2 + t| + C
= log \(\left|\frac{1+t}{2+t}\right|\) + C
= log \(\left|\frac{1+e^x}{2+e^x}\right|\) + C

(ii) I = ∫ \(\frac{d x}{e^{2 x}+1}\)
put e^x = t
⇒ e^x dx = dt
⇒ dx = \(\frac{d t}{t}\)
∴ I = ∫ \(\frac{d t}{t\left(t^2+1\right)}\)
Let \(\frac{1}{t\left(t^2+1\right)}=\frac{\mathrm{A}}{t}+\frac{\mathrm{B} t+\mathrm{C}}{t^2+1}\) ………….(1)
Multiplying both sides of eqn. (1) by t (t² + 1) ; we have
1 = A (t² + 1) + (Bt + C) t …………….(2)
putting t = 0 in eqn. (2) ; we have
1 = A
Coeff. of t² ;
0 = A + B
⇒ B = – 1
Coeff. of t ;
0 = C
∴ from (1) ;
\(\frac{1}{t\left(t^2+1\right)}=\frac{1}{t}-\frac{t}{t^2+1}\)
Thus, I = ∫ \(\left[\frac{1}{t}-\frac{t}{t^2+1}\right]\) dt

Question 14.
(i) ∫ \(\frac{\cos x}{(1-\sin x)(3-\sin x)}\) dx
(ii) ∫ \(\frac{\sin 2 x}{(1+\sin x)(2+\sin x)}\) dx (ISC 2017)
Solution:
(i) Let I = ∫ \(\frac{\cos x}{(1-\sin x)(3-\sin x)}\) dx
put sin x = t
⇒ cos x dx = dt
∴ I = ∫ \(\frac{d t}{(1-t)(3-t)}\)
Let \(\frac{1}{(1-t)(3-t)}=\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{3-t}\) ……………(1)
Multiplying both sides of eqn. (1) by (1 – t) (3 – t) ; we have
1 = A (3 – t) + B (1 – t) ………..(2)
putting t = 1, 3 successively in eqn. (2) ; we have
1 = 2A
⇒ A = \(\frac{1}{2}\)
and 1 = – 2B
⇒ B = – \(\frac{1}{2}\)

(ii) Let I = ∫ \(\frac{\sin 2 x}{(1+\sin x)(2+\sin x)}\) dx
= ∫ \(\frac{2 \sin x \cos x d x}{(1+\sin x)(2+\sin x)}\)
put sin x = t
⇒ cos x dx = dt
= ∫ \(\frac{2 t d t}{(1+t)(2+t)}\)
Let \(\frac{2 t}{(t+1)(t+2)}=\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+2}\) ………….(1)
Multiplying both sides of eqn. (1) by (t + 1) (t + 2) ; we have
2t = A (t + 2) + B (t + 1)
putting t = – 1, – 2 successively in eqn. (2) ; we get
– 2 = A
and – 4 = – B
⇒ B = 4
∴ from (1) ;
I = ∫ \(\left[\frac{-2}{t+1}+\frac{4}{t+2}\right]\) dt
= – 2 log |1 + t| + 4 log |2 + t| + C
= – 2 log |1 + sin x| + 4 log |2 + sin x| + C

Question 15.
(i) ∫ \(\frac{2 \cos x}{(1-\sin x)\left(1+\sin ^2 x\right)}\) dx
(ii) ∫ \(\frac{\sin 2 x}{\left(\sin ^2 x+1\right)\left(\sin ^2 x+3\right)}\) dx
Solution:
(i) Let I = ∫ \(\frac{2 \cos x}{(1-\sin x)\left(1+\sin ^2 x\right)}\) dx
put sin x = t
⇒ cos x dx = dt
= ∫ \(\frac{2 d t}{(1-t)\left(1+t^2\right)}\)
Let \(\frac{2}{(1-t)\left(1+t^2\right)}=\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B} t+\mathrm{C}}{1+t^2}\) …………..(1)
Multiplying both sides of eqn. (1) by (1 – t) (1 + t²) ; we get
2 = A (1 + t²) + (Bt + C) (1 – t) …………(2)
2 = 2A
⇒ A = 1
putting t = 1 in eqn. (2) ; we have
Coeff. of t² ;
0 = A – B
⇒ B = 1
Coeff. of t ;
0 = B – C
⇒ C = B = 1
∴ from eqn. (1) ; we have

(ii) Let I = ∫ \(\frac{\sin 2 x}{\left(\sin ^2 x+1\right)\left(\sin ^2 x+3\right)}\) dx
put sin² x = t
⇒ 2 sin x cos x dx = dt
⇒ sin 2x dx = dt
∴ I = ∫ \(\frac{d t}{(t+1)(t+3)}\)
Let \(\frac{1}{(t+1)(t+3)}=\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+3}\) ……………..(1)
Multiplying both sides of eqn. (1); we have
1 =A (t + 3) +B (t + 1)
putting t = – 3 in eqn. (2); we have
1 = – 2B
⇒ B = – \(\frac{1}{2}\)
putting t = – 1 in eqn. (2); we have
1 = 2A
⇒ A = \(\frac{1}{2}\)
∴ from (1); we have
I = ∫ \(\left[\frac{\frac{1}{2}}{t+1}-\frac{\frac{1}{2}}{t+3}\right]\) dt
= \(\frac{1}{2}\) log |t + 1| – \(\frac{1}{2}\) log |t + 3| + C
I = \(\frac{1}{2} \log \left|\frac{t+1}{t+3}\right|\) + C

Question 15 (old).
(ii) ∫ \(\frac{\sin 2 x}{(1-\cos 2 x)(2-\cos 2 x)}\) dx
(iii) ∫ \(\frac{\sin 2 x}{(1+\sin x)(2+\sin x)}\) dx
Solution:
(ii) Let I = ∫ \(\frac{\sin 2 x}{(1-\cos 2 x)(2-\cos 2 x)}\) dx
put cos 2x = t
⇒ – 2 sin 2x dx = dt
∴ I = – \(\frac{1}{2} \int \frac{d t}{(1-t)(2-t)}\)
Let \(\frac{1}{(1-t)(2-t)}=\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{2-t}\) ………………(1)
Multiplying eqn. (1) by (1 – t) (2 – t) ; we have
1 = A (2 – t) + B (1 – t) …………..(2)
put t = 1 in eqn. (2) ;
1 = A
put t = 2 in eqn. (2) ;
1 = – B
⇒ B = – 1
∴ From (1) ; we get
\(\frac{1}{(1-t)(2-t)}=\frac{1}{1-t}-\frac{1}{2-t}\)
∴ I = – \(\frac{1}{2}\left[\int \frac{d t}{1-t}-\int \frac{d t}{2-t}\right]\)
= – \(\frac{1}{2}\left[\frac{\log |1-t|}{-1}-\frac{\log |2-t|}{-1}\right]\) + C
= \(\frac{1}{2} \log \left|\frac{1-\cos 2 x}{2-\cos 2 x}\right|\) + C

(iii) Let I = ∫ \(\frac{\sin 2 x d x}{(1+\sin x)(2+\sin x)}\)
= ∫ \(\frac{2 \sin x \cos x d x}{(1+\sin x)(2+\sin x)}\)
put sin x = t
⇒ cos x dx = dt
= ∫ \(\frac{2 t d t}{(1+t)(2+t)}\)
Let \(\frac{2 t}{(t+1)(2+t)}=\frac{\mathrm{A}}{1+t}+\frac{\mathrm{B}}{2+t}\) ……………(1)
MuItpIying both sides of eqn. (1) by (1 + t) (2 + t) ; we get
2t = A (2 + t) + B (1 + t) ……………(2)
putting t = – 1 in eqn. (2) ; we have
⇒ – 2 = A
putting t = 2 in eqn. (2) ; we have
– 4 = – B
⇒ B = 4
∴ from (1) ;
I = ∫ \(\frac{2 t d t}{(1+t)(2+t)}\)
= \(\int \frac{-2}{1+t} d t+\int \frac{4 d t}{2+t}\)
= – 2 log |1 + t| + 4 log |2 + t| + C
= – 2 log |1 + sin x| + 4 log |2 + sin x| + C

Question 16.
(i) ∫ \(\frac{\sec ^2 x}{(2+\tan x)(3+\tan x)}\) dx
(ii) ∫ \(\frac{d x}{\sin x(3+2 \cos x)}\)
Answer:
(i) Let I = ∫ \(\frac{\sec ^2 x}{(2+\tan x)(3+\tan x)}\) dx
put tan x = t
⇒ sec² x dx = dt
∴ I = ∫ \(\frac{d t}{(2+t)(3+t)}\)
Let \(\frac{1}{(2+t)(3+t)}=\frac{\mathrm{A}}{2+t}+\frac{\mathrm{B}}{3+t}\) ………….(1)
Multiply both sides of eqn. (1) by (2 + t) (3 + t) ; we have
1 = A (3 + t) + B (2 + t)
putting t = – 2, – 3 successively in eqn. (2) ; we have
1 = A
and – 1 = B
∴ from (1) ;
\(\frac{1}{(2+t)(3+t)}=\frac{1}{2+t}-\frac{1}{3+t}\)
Thus, I = ∫ \(\left[\frac{1}{2+t}-\frac{1}{3+t}\right]\) dt
= log |2 + t| – log |3 + t|
= log \(\left|\frac{2+t}{3+t}\right|\) + C
= log \(\left|\frac{2+\tan x}{3+\tan x}\right|\) + C

(ii) Let I = ∫ \(\frac{d x}{\sin x(3+2 \cos x)}\)
= ∫ \(\frac{\sin x d x}{\left(1-\cos ^2 x\right)(3+2 \cos x)}\)
put cos x = t
– sin x dx = dt
∴ I = – ∫ \(\frac{d t}{(1-t)(1+t)(3+2 t)}\) ……………(1)
Let \(\frac{1}{(1-t)(1+t)(3+2 t)}=\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{1+t}+\frac{\mathrm{C}}{3+2 t}\) …………..(2)
Multiply both sides of eqn. (2) by (1 – t) (1 + t) (3 + 2t) ; we have
1 = A (1 + t) (3 + 2t) + B (1 – t) (3 + 2t) + C (1 – t) (1 + t)
putting t = 1, – 1, – \(\frac{3}{2}\) successively in eqn. (3) ; we get
1 = 10A
⇒ A = \(\frac{1}{10}\) ;
1 = 2B
⇒ B = \(\frac{1}{2}\)
and 1 = C \(\left(\frac{5}{2}\right)\left(\frac{-1}{2}\right)\)
⇒ C = – \(\frac{4}{5}\)
∴ from (1) ; we get
I = – \(\left[\int \frac{1 / 10}{1-t} d t+\int \frac{1 / 2 d t}{1+t}+\int \frac{-4 / 5 d t}{3+2 t}\right]\)
= – \(\left[\frac{-1}{10} \log |1-t|+\frac{1}{2} \log |1+t|-\frac{4}{5} \frac{\log |3+2 t|}{2}\right]\) + C
= \(\frac{1}{10}\) log |1 – cos x| – \(\frac{1}{2}\) log |1 + cos x| + \(\frac{2}{5}\) log |3 + 2 cos x| + C.

Question 17.
(i) ∫ \(\frac{d x}{\sin x-\sin 2 x}\)
(ii) ∫ \(\frac{d x}{x(6 \log x)^2+(7 \log x+2)}\) (ISC 2012)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sin x-\sin 2 x}\)
= ∫ \(\frac{d x}{\sin x(1-2 \cos x)}\)
put cos x = t
⇒ – sin x dx = dt
= ∫ \(\frac{\sin x d x}{\left(1-\cos ^2 x\right)(1-2 \cos x)}\)
= ∫ \(\frac{-d t}{\left(1-t^2\right)(1-2 t)}\)
Let \(\frac{-1}{\left(1-t^2\right)(1-2 t)}=\frac{-1}{(1-t)(1+t)(1-2 t)}\)
= \(\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{1+t}+\frac{\mathrm{C}}{1-2 t}\) …………..(1)
Multiplying both sides of eqn. (1) by (1 – t) (1 + t) (1 – 2t) ; we have
– 1 = A (1 + t) (1 – 2t) + B (1 – t) (1 – 2t) + C (1 – t) (1 + t) …………..(2)
putting t = 1, – 1 and \(\frac{1}{2}\) successively in eqn. (2) ; we have
– 1 = A (2) (- 1)
⇒ A = \(\frac{1}{2}\)
– 1 = B (2) (3)
⇒ B = – \(\frac{1}{6}\)
and – 1 = C \(\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\)
⇒ C = – \(\frac{4}{3}\)
∴ from eqn. (1) ; we have

(ii) Let I = ∫ \(\frac{d x}{x\left[6(\log x)^2+7 \log x+2\right]}\)
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
= ∫ \(\frac{d t}{6 t^2+7 t+2}\)
= ∫ \(\frac{d t}{(2 t+1)(3 t+2)}\)
Let \(\frac{1}{(2 t+1)(3 t+2)}=\frac{\mathrm{A}}{2 t+1}+\frac{\mathrm{B}}{3 t+2}\) …………(1)
Multiplying both sides of eqn. (1) by (2t + 1) (3t + 2) ; we have
⇒ 1 = A (3t + 2) + B (2t + 1) ……………..(2)
putting t = – \(\frac{2}{3}\) in eqn. (2) ; we have
1 = B (- \(\frac{1}{3}\))
⇒ B = – 3
putting t = – \(\frac{1}{2}\) in eqn. (2) ; we have
1 = A (latex]\frac{1}{2}[/latex])
⇒ A = 2
∴ from (1) ;
I = ∫ \(\left[\frac{2}{2 t+1}-\frac{3}{3 t+2}\right]\) dt
= \(\frac{2 \log |2 t+1|}{2}-\frac{3 \log |3 t+2|}{3}\) + C
= log |2 log x + 1| – log |3 log x + 2| + C

Question 18.
(i) ∫ \(\frac{d x}{\left(x^2+1\right)\left(x^2+4\right)}\)
(ii) ∫ \(\frac{x^2}{\left(x^2+1\right)\left(x^2+4\right)}\) dx (NCERT)
Solution:
(i) put x² = t
∴ \(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{1}{(t+1)(t+4)}=\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+4}\) …………….(1)
Multiply both sides by (t + 1) (t + 4) ; we have
1 = A (t + 4) + B (t + 1) …………..(2)
putting t = – 4, – 1 successively in eqn. (2) ; we have
1 = – 3B
⇒ B = – \(\frac{1}{3}\)
and A = \(\frac{1}{3}\)
∴ from (1) ; we have
\(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{\frac{1}{3}}{x^2+1}-\frac{\frac{1}{3}}{x^2+4}\)
Thus, \(\int \frac{d x}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{1}{3} \int \frac{1}{x^2+1^2} d x-\frac{1}{3} \int \frac{1}{x^2+2^2} d x\)
= \(\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}\) + C
[∵ \(\int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]

(ii) put x² = t
∴ \(\frac{x^2}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{t}{(t+1)(t+4)}=\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+4}\) …………(1)
Multiply both sides of eqn. (1) by (t + 1) (t + 4) ; we have
t = A (t + 4) + B (t + 1)
putting t = – 1, – 4 successively in eqn. (2) ; we have
– 1 = 3A
⇒ A = – \(\frac{1}{3}\)
and – 4 = – 3B
⇒ B = \(\frac{4}{3}\)
∴ from (1) ; we get

Question 19.
(i) ∫ \(\frac{x^2+1}{\left(x^2+4\right)\left(x^2+25\right)}\) dx
(ii) ∫ \(\frac{x^2}{x^4-x^2-12}\) dx (NCERT Exemplar)
Solution:
(i) putting x² = y
Then \(\frac{x^2+1}{\left(x^2+4\right)\left(x^2+25\right)}=\frac{y+1}{(y+4)(y+25)}\)
Let \(\frac{y+1}{(y+4)(y+25)}=\frac{\mathrm{A}}{y+4}+\frac{\mathrm{B}}{y+25}\) …………….(1)
Then y + 1 = A (y + 25) + B (y + 4) …………(2)
putting y = – 4, – 25 successively in eqn. (2) ; we have
– 3 = 21 A
⇒ A = – 1/7
and – 24 = – 21 B
⇒ B = 8/7
from (1) ; we have
\(\frac{y+1}{(y+4)(y+25)}=\frac{-1 / 7}{y+4}+\frac{8 / 7}{y+25}\)

(ii) Let I = ∫ \(\frac{x^2 d x}{x^4-x^2-12}\)
= ∫ \(\frac{x^2 d x}{\left(x^2+3\right)\left(x^2-4\right)}\)
put x² = t, we have
\(\frac{x^2}{\left(x^2+3\right)\left(x^2-4\right)}=\frac{t}{(t+3)(t-4)}\)
Let \(\frac{t}{(t+3)(t-4)}=\frac{\mathrm{A}}{t+3}+\frac{\mathrm{B}}{t-4}\) ……………..(1)
Then t = A (t – 4) + B (t + 3) …………….(2)
putting t = – 3, 4 successively in eqn. (2) ; we have
– 3 = – 7A
⇒ A = \(\frac{3}{7}\)
and 4 = 7B
⇒ B = 4/7
∴ From (1) ; we have

Question 20.
(i) ∫ \(\frac{d x}{x^4-1}\)
(ii) ∫ \(\frac{x^2}{1-x^4}\) dx (NCERT Exemplar)
(iii) ∫ \(\frac{\cos x}{\left(4+\sin ^2 x\right)\left(5-4 \cos ^2 x\right)}\) dx
Solution:
(i) Let I = ∫ \(\frac{d x}{x^4-1}\) ;
put x² = y
⇒ \(\frac{1}{x^4-1}=\frac{1}{t^2-1}\)
= \(\frac{1}{(t-1)(t+1)}\)
Let \(\) …………..(1)
⇒ 1 = A (t – 1) + B (t + 1) ………….(2)
putting t = 1, – 1 successively in eqn. (2) ; we have
1 = 2B
⇒ B = 1/2
and 1 = – 2A
⇒ A = – 1/2
∴ From (1) ;

(ii) put x² = t, we have
\(\frac{x^2}{1-x^4}=\frac{t}{1-t^2}=\frac{t}{(1-t)(1+t)}\)
Let \(\frac{t}{(1-t)(1+t)}=\frac{\mathrm{A}}{1-t}+\frac{\mathrm{B}}{1+t}\) ……………..(1)
Then t = A (1 + t) + B (1 – t)
putting t = 1, – 1 successively in eqn. (2) ; we have
1 = 2A
⇒ A = 1/2
and – 1 = 2B
⇒ B = – 1/2

(iii) Let I = ∫ \(\frac{\cos x d x}{\left(4+\sin ^2 x\right)\left(5-4 \cos ^2 x\right)}\)
= ∫ \(\frac{\cos x d x}{\left(4+\sin ^2 x\right)\left[5-4\left(1-\sin ^2 x\right)\right]}\)
= ∫ \(\frac{\cos x d x}{\left(4+\sin ^2 x\right)\left(1+4 \sin ^2 x\right)}\)
put sin x = t
⇒ cos x dx = dt
= ∫ \(\frac{d t}{\left(4+t^2\right)\left(1+4 t^2\right)}\)
put t² = y
∴ \(\frac{1}{\left(4+t^2\right)\left(1+4 t^2\right)}=\frac{1}{(4+y)(1+4 y)}\)
= \( \frac{\mathrm{A}}{4+y}+\frac{\mathrm{B}}{1+4 y}\) …………(1)
Multiplying both sides of eqn. (1) by (4 + y) (1 + 4y) ; we have
1 = A (1 + 4y) + B (4 + y) …………..(2)
putting y = – 4 in eqn. (2) ; we have
1 = A (t – 16)
⇒ A = \(-\frac{1}{15}\)
putting y = – \(\frac{1}{4}\) in eqn. (2) ; we have
1 = B (4 – \(\frac{1}{4}\))
⇒ 1 = B (\(\frac{15}{4}\))
⇒ B = \(\frac{4}{15}\)
∴ from (1) ; we have

Question 21.
(i) ∫ \(\frac{\sqrt{\cos x}}{\sin x}\) dx
(ii) ∫ \(\frac{\sin x}{\sin 4 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sqrt{\cos x}}{\sin x}\) dx
put \(\sqrt{cos x}\) = t
⇒ cos x = t²
⇒ – sin x dx = 2t dt
∴ I = ∫ \(\frac{t \times 2 t d t}{-\left(1-t^4\right)}\)
= – 2 ∫ \(\frac{t^2}{\left(1-t^2\right)\left(1+t^2\right)}\) dt ……………..(1)
put t² = y
∴ \(\frac{t^2}{\left(1-t^2\right)\left(1+t^2\right)}=\frac{y}{(1-y)(1+y)}\)
= \(\frac{\mathrm{A}}{1-y}+\frac{\mathrm{B}}{1+y}\) ……………..(2)
Multiply both sides of eqn. (1) by (1 – y) (1 + y) ; we have
∴ y = A (1 + y) + B (1 – y) …………………(3)
putting y = 1, – 1 successively in eqn. (3) ; we have
1 = 2A
A = \(\frac{1}{2}\) and B = – \(\frac{1}{2}\)
∴ from (2); we get

(ii) Let I = ∫ \(\frac{\sin x}{\sin 4 x}\) dx
= ∫ \(\frac{\sin x d x}{2 \sin 2 x \cos 2 x}\) dx
= ∫ \(\frac{d x}{4 \cos x\left(1-2 \sin ^2 x\right)}\) dx
put sin x = t
⇒ cos x dx = dt = ∫ \(\frac{d t}{4\left(1-t^2\right)\left(1-2 t^2\right)}\) ………………(1)
put t² = y
∴ \(\frac{1}{\left(1-t^2\right)\left(1-2 t^2\right)}=\frac{1}{(1-y)(1-2 y)}=\frac{\mathrm{A}}{1-y}+\frac{\mathrm{B}}{1-2 y}\) ………………(2)
Multiply both sides of eqn. (2) by (1 – y) (1 – 2y) ; we get
1 = A(1 – 2y) + B (1 – y)
putting y = 1, \(\frac{1}{2}\) successively in eqn. (1) ; we have
1 = – A
⇒ A = – 1 and B = 2
∴ from (2) ; we have
\(\frac{1}{\left(1-t^2\right)\left(1-2 t^2\right)}=\frac{-1}{1-t^2}+\frac{2}{1-2 t^2}\)
∴ from (1) ; we get

Question 22.
(i) ∫ \(\frac{1+x^2}{1+x^4}\) dx
(ii) ∫ \(\frac{1-x^2}{1+x^4}\) dx
Solution:
(i) Let I = ∫ \(\frac{1+x^2}{1+x^4}\) dx
Divide Num. and Deno. by x² ; we get
∴ I = ∫ \(\frac{\left(1+\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}}\) dx
Put x – \(\frac{1}{x}\) = t
(1 + \(\frac{1}{x^2}\)) dx = dt ;
On Squaring ;
x² + \(\frac{1}{x^2}\) = t² + 2
∴ I = ∫ \(\frac{d t}{t^2+2}\)
= ∫ \(\frac{d t}{t^2+(\sqrt{2})^2}\)
[using \(\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]
∴ I = \(\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{t}{\sqrt{2}}\right)\) + C
= \(\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{x^2-1}{\sqrt{2} x}\right)\) + C

(ii) Let I = ∫ \(\frac{1-x^2}{1+x^4}\) dx
Divide Num. and Deno. by x² ; we get

Question 23.
(i) ∫ \(\frac{x^2-1}{x^4+x^2+1}\) dx
(ii) ∫ \(\frac{x^2+4}{x^4+x^2+16}\) dx
Solution:
(i) Let I = ∫ \(\frac{x^2-1}{x^4+x^2+1}\) dx
Divide Num. and Deno. by x² ; we get
= ∫ \(\frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}\) dx
put x + \(\frac{1}{x}\) = t
⇒ (1 – \(\frac{1}{x^2}\)) dx = dt
On squaring; we get
(x + \(\frac{1}{x}\))² = t²
⇒ x² + \(\frac{1}{x^2}\) = t² – 2

(ii) Let I = ∫ \(\frac{x^2+4}{x^4+x^2+16}\) dx
Divide Numerator and denominator by x² ; we have
I = ∫ \(\frac{\left(1+\frac{4}{x^2}\right) d x}{x^2+1+\frac{16}{x^2}}\)
put x – \(\frac{4}{x}\) = t
⇒ (1 + \(\frac{4}{x^2}\)) dx = dt
On squaring; we have
(x – \(\frac{4}{x}\))² = t²
⇒ x² + \(\frac{16}{x^2}\) – 8 = t²
⇒ x² + \(\frac{16}{x^2}\) = t² + 8
∴ I = ∫ \(\frac{d t}{t^2+8+1}\)
= \(\frac{d t}{t^2+3^2}\)
= \(\frac{1}{3} \tan ^{-1}\left(\frac{t}{3}\right)\) + C
= \(\frac{1}{3} \tan ^{-1}\left(\frac{x^2-4}{3 x}\right)\) + C

Question 24.
(i) ∫ \(\frac{1}{x^4+1}\) dx
(ii) ∫ \(\frac{x^2}{x^4+1}\) dx
Solution:
(i) ∫ \(\frac{1}{x^4+1}\) dx
= \(\frac{1}{2} \int \frac{2 d x}{x^4+1}\)
= \(\frac{1}{2} \int \frac{\left[\left(x^2+1\right)-\left(x^2-1\right)\right]}{x^4+1}\) dx

(ii) Let I = ∫ \(\frac{x^2}{x^4+1}\) dx
∴ I = \(\frac{1}{2} \int \frac{\left(x^2+1\right)+\left(x^2-1\right)}{x^4+1}\)
= \(\frac{1}{2} \int \frac{x^2+1}{x^4+1} d x+\frac{1}{2} \int \frac{x^2-1}{x^4+1} d x\)
= \(\frac{1}{2}\) I₁ – \(\frac{1}{2}\) I₂ ……………..(1)
where I₁ = ∫ \(\frac{x^2}{x^4+1}\) dx
Divide Num. and Deno. by x² ; we get

Question 25.
(i) ∫ \(\frac{x^2}{x^4+x^2+1}\) dx
(ii) ∫ \(\frac{x^2-3 x+1}{x^4-x^2+1}\) dx
Solution:
(i) Let I = ∫ \(\frac{x^2}{x^4+x^2+1}\) dx

(ii) Let I = ∫ \(\frac{x^2-3 x+1}{x^4-x^2+1}\) dx
∴ I = \(\int \frac{\left(x^2+1\right) d x}{x^4-x^2+1}-3 \int \frac{x d x}{x^4-x^2+1}\)
⇒ I = I₁ – 3 I₂ …………….(1)
where I₁ = ∫ \(\frac{\left(x^2+1\right) d x}{x^4-x^2+1}\)
Divide Num. and Deno. by x² ; we have
= ∫ \(\frac{\left(1+\frac{1}{x^2}\right) d x}{x^2-1+\frac{1}{x^2}}\)
put x – \(\frac{1}{x}\) = t
⇒ (1 + \(\frac{1}{x^2}\)) dx = dt
On squaring ;
(x – \(\frac{1}{x}\))² = t²
⇒ x² + \(\frac{1}{x^2}\) – 2 = t²
⇒ x² + \(\frac{1}{x^2}\) = t² + 2

Question 26.
(i) ∫ \(\frac{\sin x+\cos x}{\sin ^2 x+\cos ^4 x}\) dx
(ii) ∫ (\(\sqrt{\cot x}+\frac{1}{\sqrt{\cot x}}\)) dx
Solution:
(i) Let

put sin x – cos x = t
⇒ (cos x + sin x) dx = dt
On squaring ; we have
(sin x – cos x)² = t²
⇒ sin² x + cos² x – sin2x = t²
⇒ 1 – sin 2x = t²
⇒ sin 2x = 1 – t²
∴ I = 4 ∫ \(\frac{d t}{4-\left(1-t^2\right)^2}\)
= ∫ \(\frac{4 d t}{\left[2-\left(1-t^2\right)\right]\left[2+1-t^2\right]}\)
= 4 ∫ \(\frac{d t}{\left(1+t^2\right)\left(3-t^2\right)}\) ……………….(1)
put t² = y
∴ \(\frac{1}{\left(1+t^2\right)\left(3-t^2\right)}=\frac{1}{(1+y)(3-y)}=\frac{\mathrm{A}}{1+y}+\frac{\mathrm{B}}{3-y}\) ……………(2)
Multiplying both sides by (1 + y) (3 – y) ; we have
1 = A (3 – y) + B (1 + y) …………(3)
putting y = – 1, 3 successively in eqn. (3) ; we have
1 = 4A
⇒ A = \(\frac{1}{4}\)
and B = \(\frac{1}{4}\)
∴ from (2) ; we have

(ii) Let I = ∫ (\(\sqrt{\cot x}+\frac{1}{\sqrt{\cot x}}\)) dx
= ∫ [latex]\sqrt{\tan x}+\frac{1}{\sqrt{\tan x}}[/latex] dx
put \(\sqrt{\tan x}\) = t
⇒ tan x = t²
sec² dx = 2t dt
⇒ dx = \(\frac{2 t d t}{1+\tan ^2 x}=\frac{2 t d t}{1+t^4}\)
= ∫ \(\left(t+\frac{1}{t}\right) \frac{2 t d t}{1+t^4}\)
= ∫ \(\frac{2\left(t^2+1\right) d t}{t^4+1}\)
Divide Numerator and denominator by t² ; we have
= 2 ∫ \(\frac{\left(1+\frac{1}{t^2}\right) d t}{t^2+\frac{1}{t^2}}\)
put t – \(\frac{1}{t}\) = u
⇒ (1 + \(\frac{1}{t^2}\)) dt = du
On squaring; we have

Question 26 (0ld).
(ii) ∫ \(\sqrt{cot x}\) dx
Solution:
Let I = ∫ \(\sqrt{cot x}\) dx
put \(\sqrt{cot x}\) = t
⇒ cot x = t²
⇒ – cosec² x dx = 2t dt

Well-structured Understanding ISC Mathematics Class 12 Solutions Chapter 8 Integrals Ex 8.8 facilitate a deeper understanding of mathematical principles.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.8

Evaluate the following (1 to 12) integrals:

Question 1.
(i) ∫ \(\frac{d x}{x^2-4 x+8}\)
(ii) ∫ \(\frac{d x}{x^2+2 x+2}\)
Solution:
(i) ∫ \(\frac{d x}{x^2-4 x+8}\)
= ∫ \(\frac{d x}{x^2-4 x+4+4}\)
= ∫ \(\frac{d x}{(x-2)^2+2^2}\)
[put x – 2 = t
⇒ dx = dt]
= ∫ \(\frac{d t}{t^2+a^2}\)
= \(\frac{1}{2} \tan ^{-1} \frac{t}{2}\) + C
[∵ ∫ \(\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\)]
= \(\frac{1}{2} \tan ^{-1}\left(\frac{x-2}{2}\right)\) + C

(ii) ∫ \(\frac{d x}{x^2+2 x+2}\)
Solution:
Let I = ∫ \(\frac{d x}{x^2+2 x+2}\)
= ∫ \(\frac{d x}{x^2+2 x+1+1}\)
= ∫ \(\frac{d x}{(x+1)^2+1^2}\)
= ∫ \(\frac{d t}{t^2+1^2}\) ;
when x + 1 = t
⇒ dx = dt
= \(\frac{1}{1} \tan ^{-1}\left(\frac{t}{1}\right)\) + C
= tan^-1 (x + 1) + C

Question 1 (old).
(ii) ∫ \(\frac{d x}{5-8 x-x^2}\) (NCERT)
Solution:
Let I = ∫ \(\frac{d x}{5-8 x-x^2}\)

Question 2.
(i) ∫ \(\frac{d x}{x^2-6 x+13}\) (NCERT)
(ii) ∫ \(\frac{x+3}{x^2-2 x-5}\) dx (NCERT)
Solution:
(i) ∫ \(\frac{d x}{x^2-6 x+13}\)
= ∫ \(\frac{d x}{x^2-6 x+9+4}\)
= ∫ \(\frac{d x}{(x-3)^2+2^2}\)
put x – 3 = t
dx = dt
= ∫ \(\frac{d t}{t^2+2^2}\)
= \(\frac{1}{2} \tan ^{-1} \frac{t}{2}\) + C
= \(\frac{1}{2} \tan ^{-1}\left(\frac{x-3}{2}\right)\) + C

(ii) Let I = ∫ \(\frac{x+3}{x^2-2 x-5}\) dx

Question 3.
(i) ∫ \(\frac{x+2}{2 x^2+6 x+5}\) dx
(ii) ∫ \(\frac{2 x+1}{18-4 x-x^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{x+2}{2 x^2+6 x+5}\) dx

(ii) Let I = ∫ \(\frac{2 x+1}{18-4 x-x^2}\) dx
Let 2x + 1 = A \(\frac{d}{d x}\) (18 – 4x – x²) + B
= A (- 4 – 2x) + B
On comparing the coefficient of x and constant terms on both sides ; we get
2 = – 2 A
⇒ A = – 1 ;
1 = – 4A + B
B = – 3

Question 4.
(i) ∫ \(\frac{e^x}{2 e^{2 x}+3 e^x+5}\) dx
(ii) ∫ \(\frac{\cos x}{6+4 \sin x-\cos ^2 x}\) dx
Solution:
(i) ∫ \(\frac{e^x}{2 e^{2 x}+3 e^x+5}\) dx ;
put e^x = t
e^x dx = dt
∴ I = ∫ \(\frac{d t}{2 t^2+3 t+5}\)
= \(\frac{1}{2}\) ∫ \(\frac{d t}{t^2+\frac{3 t}{2}+\frac{5}{2}}\)
= \(\frac{1}{2}\) ∫ \(\frac{d t}{t^2+\frac{3 t}{2}+\frac{9}{16}-\frac{9}{16}+\frac{5}{2}}\)

(ii) Let I = ∫ \(\frac{\cos x}{6+4 \sin x-\cos ^2 x}\) dx

= tan^-1 (t + 2) + C
= tan^-1 (2 + sin x) + C

Question 5.
(i) ∫ \(\frac{d x}{\sqrt{2 x-x^2}}\) (NCERT)
(ii) ∫ \(\frac{d x}{\sqrt{7-6 x-x^2}}\)
Solution:
(i) ∫ \(\frac{d x}{\sqrt{2 x-x^2}}\)
= ∫ \(\frac{d x}{\sqrt{-\left(x^2-2 x+1-1\right)}}\)
= ∫ \(\frac{d x}{\sqrt{1-(x-1)^2}}\)
put x – 1 = t
⇒ dx = dt
= ∫ \(\frac{d t}{\sqrt{1-t^2}}\)
= sin^-1 t + C
= sin^-1 (x – 1) + C
[∵ ∫ \(\frac{d x}{\sqrt{a^2-x^2}}\) = sin^-1 \(\frac{x}{a}\) + C]

(ii) Let I = ∫ \(\frac{d x}{\sqrt{7-6 x-x^2}}\)

Question 5 (old).
(ii) ∫ \(\frac{d x}{\sqrt{7-6 x-x^2}}\)
Solution:
Let I = ∫ \(\frac{d x}{\sqrt{7-6 x-x^2}}\)
= ∫ \(\frac{d x}{\sqrt{x^2-3 x+\frac{9}{4}-\frac{9}{4}+2}}\)
= ∫ \(\frac{d x}{\sqrt{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2}}\)
put x – \(\frac{3}{2}\) = t
⇒ dx = dt
= ∫ \(\frac{d t}{\sqrt{t^2-\left(\frac{1}{2}\right)^2}}\)
= log |t + \(\sqrt{t^2-\frac{1}{4}}\)| + C
= log |x – \(\frac{3}{2}+\sqrt{x^2-3 x+2}\)| + C
[∵∫ \(\frac{d x}{\sqrt{x^2-a^2}}\) = log |x + \(\sqrt{x^2-a^2}\)| + C|]

Question 6.
(i) ∫ \(\frac{d x}{\sqrt{7-6 x-x^2}}\) (NCERT)
(ii) ∫ \(\frac{d x}{\sqrt{(x-1)(x-2)}}\) (NCERT)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{7-6 x-x^2}}\)

(ii) Let I = ∫ \(\frac{d x}{\sqrt{(x-1)(x-2)}}\)

Question 6 (old).
(ii) ∫ \(\frac{d x}{\sqrt{x^2+2 x+2}}\) (NCERT)
Solution:
Let I = ∫ \(\frac{d x}{\sqrt{x^2+2 x+2}}\)
= ∫ \(\frac{d x}{\sqrt{x^2+2 x+1+1}}\)
= ∫ \(\frac{d x}{\sqrt{(x+1)^2+1^2}}\)
put x + 1 = t
⇒ dx = dt
= ∫ \(\frac{d t}{\sqrt{t^2+1^2}}\)
= log |t + \(\sqrt{t^2+1}\)| + C
[∵∫ \(\frac{d x}{\sqrt{x^2-a^2}}\) = log |x + \(\sqrt{x^2-a^2}\)| + C|]
= log |x + 1 + \(\sqrt{x^2+2 x+2}\)| + C

Question 7.
(i) ∫ \(\frac{d x}{\sqrt{5-4 x-2 x^2}}\)
(ii) ∫ \(\frac{d x}{\sqrt{5 x-4 x^2}}\) (ISC 2020)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{5-4 x-2 x^2}}\)
= ∫ \(\frac{d x}{\sqrt{-2\left(x^2+2 x-\frac{5}{2}\right)}}\)
= ∫ \(\frac{d x}{\sqrt{2} \sqrt{-\left(x^2+2 x+1-1-\frac{5}{2}\right)}}\)
= \(\frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{-(x+1)^2+\frac{7}{2}}}\)
= \(\frac{1}{\sqrt{2}} \int \frac{d x}{\sqrt{\left(\sqrt{\frac{7}{2}}\right)^2-(x+1)^2}}\)
= \(\frac{1}{\sqrt{2}} \sin ^{-1}\left(\frac{x+1}{\sqrt{\frac{7}{2}}}\right)\) + c
= \(\frac{1}{\sqrt{2}} \sin ^{-1}\left[\sqrt{\frac{2}{7}}(x+1)\right]\) + c

(ii) Let I = ∫ \(\frac{d x}{\sqrt{5 x-4 x^2}}\)

Question 7 (old).
(ii) ∫ \(\frac{d x}{\sqrt{8+3 x-x^2}}\)
Solution:
Let I = ∫ \(\frac{d x}{\sqrt{8+3 x-x^2}}\)
= ∫ \(\frac{d x}{\sqrt{-\left(x^2-3 x-8\right)}}\)
= ∫ \(\frac{d x}{\sqrt{-\left(x^2-3 x+\frac{9}{4}-\frac{9}{4}-8\right)}}\)
= ∫ \(\frac{d x}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^2}}\)
= ∫ \(\frac{d x}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}}\)
= sin^-1 \(\left(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\right)\) + c
= sin^-1 \(\left(\frac{2 x-3}{\sqrt{41}}\right)\) + c
[∵∫ \(\frac{d x}{\sqrt{a^2-x^2}}\) = sin^-1 \(\frac{x}{a}\) + c]

Question 8.
(i) ∫ \(\frac{x-1}{\sqrt{x^2-x}}\) dx (NCERT)
(ii) ∫ \(\frac{5 x+3}{\sqrt{x^2+4 x+10}}\) dx
Solution:
(i) Let x – 1 = A \(\frac{d}{d x}\) (x² – x) + B
= A (2x – 1) + B
⇒ x – 1 = 2Ax – A + B
∴ 2A = 1
A = \(\frac{1}{2}\)
and – A + B = – 1
⇒ B = – 1 + \(\frac{1}{2}\) = – \(\frac{1}{2}\)
∴ I = ∫ \(\frac{x-1}{\sqrt{x^2-x}}\) dx
= ∫ \(\frac{\frac{1}{2}(2 x-1)-\frac{1}{2}}{\sqrt{x^2-x}}\) dx
= \(\frac{1}{2} \int \frac{(2 x-1) d x}{\sqrt{x^2-x}}-\frac{1}{2} \int \frac{d x}{\sqrt{x^2-x}}\)
= \(\frac{1}{2}\) I₁ – \(\frac{1}{2}\) I₂
where I₁ = \(\frac{(2 x-1) d x}{\sqrt{x^2-x}}\)
put x² – x = t
⇒ (2x – 1) dx = dt

(ii) Let I = ∫ \(\frac{5 x+3}{\sqrt{x^2+4 x+10}}\) dx

Question 8 (old).
(i) ∫ \(\frac{d x}{\sqrt{5 x^2-2 x}}\) (NCERT)
(ii) ∫ \(\frac{d x}{\sqrt{9 x-4 x^2}}\) (NCERT)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{5 x^2-2 x}}\)
= ∫ \(\frac{d x}{\sqrt{5\left(x^2-\frac{2}{5} x+\frac{1}{25}-\frac{1}{25}\right)}}\)
= \(\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2}}\)
= \(\frac{1}{\sqrt{5}} \log \left|x-\frac{1}{5}+\sqrt{\left(x-\frac{1}{5}\right)^2-\left(\frac{1}{5}\right)^2}\right|\) + c
= \(\frac{1}{\sqrt{5}} \log \left|\frac{5 x-1}{5}+\frac{\sqrt{5 x^2-2 x}}{\sqrt{5}}\right|\) + c
[∵∫ \(\frac{d x}{\sqrt{a^2-x^2}}\) = log |x + \(\sqrt{x^2-a^2}\)| + c]

(ii) Let I = ∫ \(\frac{d x}{\sqrt{9 x-4 x^2}}\)

Question 9.
(i) ∫ \(\frac{3 x+5}{\sqrt{x^2-8 x+7}}\) dx
(ii) ∫ \(\frac{x^2}{\sqrt{x^6-2 x^2-3}}\) dx
Solution:
(i) Let I = ∫ \(\frac{3 x+5}{\sqrt{x^2-8 x+7}}\) dx

∴ from (1) ; we have
I = 3 \(\sqrt{x^2-8 x+7}\) + 17 log |x – 4 + \(\sqrt{x^2-8 x+7}\)| + C

(ii) Let I = ∫ \(\frac{x^2}{\sqrt{x^6-2 x^2-3}}\) dx ;
put x³ = t
⇒ 3x² dx = dt
∴ I = ∫ \(\frac{d t}{3 \sqrt{t^2-2 t-3}}\)
= \(\frac{1}{3} \int \frac{d t}{\sqrt{t^2-2 t+1-4}}\)
= \(\frac{1}{3} \int \frac{d t}{\sqrt{(t-1)^2-2^2}}\)
put t – 1 = u
⇒ dt = du
= \(\frac{1}{3} \int \frac{d u}{\sqrt{u^2-2^2}}\)
= \(\frac{1}{3}\) log |u + \(\sqrt{u^2-4}\)| + C
[∵∫ \(\frac{d x}{\sqrt{x^2+^2}}\) = log |x + \(\sqrt{x^2-a^2}\)| + c]
= \(\frac{1}{3}\) log |t – 1 + \(\sqrt{t^2-2 t-3}\)|
= \(\frac{1}{3}\) log |x – 1 + \(\sqrt{x^6-2 x^3-3}\)| + C

Question 10 (old).
(i) ∫ \(\frac{x+3}{\sqrt{5-4 x-x^2}}\) dx
Solution:
Let I = \(\frac{(x+3) d x}{\sqrt{5-4 x-x^2}}\)

Question 11 (old).
(ii) ∫ \(\frac{x+2}{\sqrt{(x-2)(x-3)}}\) dx
Solution:
Let I = ∫ \(\frac{x+2}{\sqrt{(x-2)(x-3)}}\) dx

Question 12 (old).
(i) ∫ \(\frac{x+2}{\sqrt{x^2+2 x+3}}\) dx
Solution:
Let I = ∫ \(\frac{x+2}{\sqrt{x^2+2 x+3}}\) dx

Students can track their progress and improvement through regular use of ML Aggarwal Maths for Class 12 Solutions Chapter 8 Integrals Ex 8.7.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.7

Very Short answer type questions (1 to 3):

Evaluate the following (1 to 16) integrals:

Question 1.
(i) ∫ \(\frac{1}{x^2-16}\) dx (NCERT)
(ii) ∫ \(\frac{1}{9-x^2}\) dx
(iii) ∫ \(\frac{d x}{\sqrt{1-x^2}}\)
(iv) ∫ \(\frac{d x}{x^2+16}\)
Solution:
(i) ∫ \(\frac{1}{x^2-16}\) dx
= (i) ∫ \(\frac{d x}{x^2-16}\)
= \(\frac{1}{2 \times 4} \log \left|\frac{x-4}{x+4}\right|\) + C
= \(\frac{1}{8} \log \left|\frac{x-4}{x+4}\right|\) + C
[∵ ∫ \(\frac{d x}{x^2-a^2}\) = \(\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\) + C]

(ii) ∫ \(\frac{1}{9-x^2}\) dx
= ∫ \(\frac{d x}{3^2-x^2}\)
= \(\frac{1}{2 \times 3} \log \left|\frac{3+x}{3-x}\right|\) + C
= \(\frac{1}{6} \log \left|\frac{3+x}{3-x}\right|\) + C
[∵ ∫ \(\frac{d x}{x^2-a^2}\) = \(\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|\) + C]

(iii) ∫ \(\frac{d x}{\sqrt{1-x^2}}\)
= ∫ \(\frac{d x}{\sqrt{1^2-x^2}}\)
= sin^-1 \(\frac{x}{1}\) + C
= sin^-1 x + C
[∵ ∫ \(\frac{d x}{a^2-x^2}\) = sin^-1 \(\frac{x}{a}\) + C]

(iv) ∫ \(\frac{d x}{x^2+16}\)
= ∫ \(\frac{d x}{x^2+4^2}\)
=\(\frac{1}{4}\) tan^-1 \(\frac{x}{4}\) + C
[∵ ∫ \(\frac{d x}{x^2+a^2}\) = \(\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]

Question 2.
(i) ∫ \(\frac{d x}{\sqrt{9-x^2}}\)
(ii) ∫ \(\frac{d x}{\sqrt{25+9 x^2}}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{9-x^2}}\)
= \(\frac{1}{2} \int \frac{d x}{\sqrt{\left(\frac{3}{2}\right)^2-x^2}}\)
= \(\frac{1}{2} \sin ^{-1}\left(\frac{x}{\frac{3}{2}}\right)\) + C
[∵ ∫ \(\frac{d x}{x^2-a^2}\) = sin^-1 \(\frac{x}{a}\) + C]
= \(\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{3}\right)\) + C

(ii) ∫ \(\frac{d x}{\sqrt{25+9 x^2}}\)
= \(\frac{1}{9} \int \frac{d x}{x^2+\left(\frac{5}{3}\right)^2}\)
= \(\frac{1}{9} \times \frac{1}{\frac{5}{3}} \tan ^{-1}\left(\frac{x}{\frac{5}{3}}\right)\) + C
= \(\frac{1}{15} \tan ^{-1}\left(\frac{3 x}{5}\right)\) + C
[∵ ∫ \(\frac{d x}{x^2+a^2}\) = \(\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]

Question 2 (old).
(i) ∫ \(\frac{d x}{\sqrt{16-9 x^2}}\) (NCERT Exemplar)
Solution:
(i) ∫ \(\frac{d x}{\sqrt{16-9 x^2}}\)
= \(\frac{1}{3} \int \frac{d x}{\sqrt{\left(\frac{4}{3}\right)^2-x^2}}\)
= \(\frac{1}{3} \sin ^{-1}\left(\frac{x}{\frac{4}{3}}\right)\) + C
[∵ ∫ \(\frac{d x}{a^2-x^2}\) = sin^-1 \(\frac{x}{a}\) + C]
= \(\frac{1}{3} \sin ^{-1}\left(\frac{3 x}{4}\right)\) + C

Question 3.
(i) ∫ \(\frac{d x}{x \sqrt{9 x^2-16}}\)
(ii) ∫ \(\frac{d x}{\sqrt{x^2+5}}\)
Solution:
(i) ∫ \(\frac{d x}{x \sqrt{9 x^2-16}}\)
= \(\frac{1}{3} \int \frac{d x}{x \sqrt{x^2-\left(\frac{4}{3}\right)^2}}\)
= \(\frac{1}{3} \times \frac{1}{\frac{4}{3}} \sec ^{-1}\left(\frac{x}{\frac{4}{3}}\right)\) + C
= \(\frac{1}{4} \sec ^{-1}\left(\frac{3 x}{4}\right)\) + C
[∵ ∫ \(\frac{d x}{x \sqrt{x^2-a^2}}=\frac{1}{a} \sec ^{-1} \frac{x}{a}\) + C]

(ii) Let I = ∫ \(\frac{d x}{\sqrt{x^2+5}}\)
= ∫ \(\frac{d x}{\sqrt{x^2+(\sqrt{5})^2}}\)
= log |x + \(\sqrt{x^2+5}\)| + C
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + C

Question 3 (old).
(ii) ∫ \(\frac{d x}{\sqrt{x^2+4}}\)
Solution:
∫ \(\frac{d x}{\sqrt{x^2+4}}\)
= ∫ \(\frac{d x}{\sqrt{x^2+2^2}}\)
= log |x + \(\sqrt{x^2+4}\)| + C
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + C

Question 4.
(i) ∫ \(\frac{1}{9 x^2-1}\) dx
(ii) ∫ \(\frac{d x}{32-2 x^2}\)
Solution:
(i) ∫ \(\frac{1}{9 x^2-1}\) dx
= \(\frac{1}{9} \int \frac{d x}{x^2-\left(\frac{1}{3}\right)^2}\)
= \(\frac{1}{9} \times \frac{1}{2 \times \frac{1}{3}} \log \left|\frac{x-\frac{1}{3}}{x+\frac{1}{3}}\right|\) + C
= \(\frac{1}{6} \log \left|\frac{3 x-1}{3 x+1}\right|\) + C

(ii) ∫ \(\frac{d x}{32-2 x^2}\)
= \(\frac{1}{2} \int \frac{d x}{16-x^2}\)
= \(\frac{1}{2} \int \frac{d x}{4^2-x^2}\)
= \(\frac{1}{2} \times \frac{1}{2 \times 4} \log \left|\frac{4+x}{4-x}\right|\) + C
= \(\frac{1}{16} \log \left|\frac{4+x}{4-x}\right|\) + C
[∵ ∫ \(\frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\) + C]

Question 5.
(i) ∫ \(\frac{d x}{\sqrt{15-8 x^2}}\)
(ii) ∫ \(\frac{d x}{\sqrt{1+4 x^2}}\) (NCERT)
Solution:
(i) ∫ \(\frac{d x}{\sqrt{15-8 x^2}}\)

(ii) ∫ \(\frac{d x}{\sqrt{1+4 x^2}}\)
= \(\frac{1}{2} \int \frac{d x}{\sqrt{\left(\frac{1}{2}\right)^2+x^2}}\)
= \(\frac{1}{2} \log \left|x+\sqrt{x^2+\frac{1}{4}}\right|\) + C
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + C

Question 6.
(i) ∫ \(\frac{2 y^2}{y^2+4}\) dy (ISC 2015)
(ii) ∫ \(\frac{x^4+1}{x^2+1}\) dx
Solution:
(i) ∫ \(\frac{2 y^2}{y^2+4}\) dy
= 2 ∫ \(\frac{y^2+4-4}{y^2+4}\) dy
= 2 ∫ [1 – \(\frac{4}{y^2+4}\)] dy
= 2 ∫ dy – 8 ∫ \(\frac{d y}{y^2+2^2}\)
= 2y – \(\frac{8}{2} \tan ^{-1}\left(\frac{y}{2}\right)\) + C
= 2y – 4 tan^-1 (\(\frac{y}{2}\)) + C

(ii) Let I = ∫ \(\frac{x^4+1}{x^2+1}\) dx
= ∫ [x² – 1 + \(\frac{2}{x^2+1}\)] dx
= \(\frac{x^{3}}{3}\) – x + 2 tan^-1 \(\frac{x}{1}\) + c
= \(\frac{x^{3}}{3}\) – x + 2 tan^-1 x + c
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + c]

Question 7.
(i) ∫ \(\frac{x}{x^4-1}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{3 x^2}{1+x^6}\) dx (NCERT)
Solution:
(i) I = ∫ \(\frac{x}{x^4-1}\) dx
put x² = t
⇒ 2x dx = dt
∴ I = ∫ \(\int \frac{d t}{2\left(t^2-1\right)}\)
= \(\frac{1}{2} \int \frac{d t}{t^2-1^2}\)
= \(\frac{1}{2} \times \frac{1}{2 \times 1} \log \left|\frac{t-1}{t+1}\right|\) + C
= \(\frac{1}{4} \log \left|\frac{x^2-1}{x^2+1}\right|\) + C
[∵ ∫ \(\frac{d x}{x^2-a^2}=\frac{1}{2 a} \log \mid \frac{x-a}{x+a}\) + C]

(ii) Let I = \(\frac{3 x^2}{1+x^6}\) dx
∴ I = ∫ \(\frac{d t}{t^2+1}\)
= tan^-1 t + C
= tan^-1 (x³) + C

Question 8.
(i) ∫ \(\frac{x^2}{\sqrt{x^6+a^6}}\) dx (NCERT)
(ii) ∫ \(\frac{x^3}{\sqrt{1-x^8}}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{x^2}{\sqrt{x^6+a^6}}\) dx
put x³ = t
⇒ 3x² dx = dt
∴ I = ∫ \(\frac{d t}{3 \sqrt{t^2+\left(a^3\right)^2}}\)
= \(\frac{1}{3}\) log |t + \(\sqrt{t^2+a^6}\)| + C
= \(\frac{1}{3}\) log |x³ + \(\sqrt{x^6+a^6}\)| + C

(ii) I = ∫ \(\frac{x^3}{\sqrt{1-x^8}}\) dx
put x⁴ = t
⇒ 4x³ dx = dt
∴ I = \(\int \frac{d t}{4 \sqrt{1-t^2}}\)
= \(\frac{1}{4} \sin ^{-1} \frac{t}{1}\) + C
= \(\frac{1}{4}\) sin^-1 (x⁴) + C
[∵ ∫ \(\frac{d x}{\sqrt{a^2-x^2}}\) = sin^-1 \(\frac{x}{a}\) + C]

Question 9.
(i) ∫ \(\frac{\sqrt{x}}{\sqrt{a^3-x^3}}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{3 x}{1+2 x^4}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{\sqrt{x} d x}{\sqrt{a^3-x^3}}\)

(ii) Let I = ∫ \(\frac{3 x}{1+2 x^4}\) dx
= ∫ \(\frac{3 x}{1+2\left(x^2\right)^2}\) dx
put x² = t
∴ 2x dx = dt
∴ I = 3 ∫ \(\frac{d t}{2\left[1+2 t^2\right]}\)
= \(\frac{3}{4} \int \frac{d t}{t^2+\frac{1}{2}}\)
= \(\frac{3}{4} \int \frac{d t}{t^2+\left(\frac{1}{\sqrt{2}}\right)^2}\)
= \(\frac{3}{4} \cdot \frac{1}{\frac{1}{\sqrt{2}}} \tan ^{-1}\left(\frac{t}{\frac{1}{\sqrt{2}}}\right)\) + C
= \(\frac{3 \sqrt{2}}{4}\) tan^-1 (√2 x²) + C

Question 10.
(i) ∫ \(\frac{\cos x}{\sqrt{4-\sin ^2 x}}\) dx (NCERT)
(ii) ∫ \(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{\cos x}{\sqrt{4-\sin ^2 x}}\) dx
put sin x = t
⇒ cos x dx = dt
= ∫ \(\frac{d t}{\sqrt{4-t^2}}\)
= ∫ \(\frac{d t}{\sqrt{2^2-t^2}}\)
= sin^-1 (\(\frac{t}{2}\)) + c
= sin^-1 (\(\left(\frac{\sin x}{2}\right)\)) + c

(ii) Let I = ∫ \(\frac{\sec ^2 x}{\sqrt{\tan ^2 x+4}}\) dx
put tan x = t
⇒ ∫ sec² x dx = dt
= ∫ \(\frac{d t}{\sqrt{4+t^2}}\)
= ∫ \(\frac{d t}{\sqrt{t^2+2^2}}\)
= log |t + \(\sqrt{t^2+4}\)| + c
[∵ ∫ \(\frac{d x}{\sqrt{x^2+a^2}}\) = log |x + \(\sqrt{x^2+a^2}\)| + c
= log |tan x + \(\sqrt{\tan ^2 x+4}\)| + c

Question 11.
(i) ∫ \(\frac{\sin x}{16-9 \cos ^2 x}\) dx
(ii) ∫ \(\frac{\tan ^2 x \sec ^2 x}{1-\tan ^6 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin x}{16-9 \cos ^2 x}\) dx
put cos x = t
⇒ – sin x dx = dt

(ii) Let I = ∫ \(\frac{\tan ^2 x \sec ^2 x}{1-\tan ^6 x}\) dx
put tan³ x = t
⇒ 3 tan² x sec² x dx = dt
∴ I = ∫ \(\frac{d t}{3\left(1-t^2\right)}\)
= \(\frac{1}{3} \times \frac{1}{2 \times 1} \log \left|\frac{1+t}{1-t}\right|\) + C
[∵ \(\frac{d x}{a^2+x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\) + C]
= \(\frac{1}{6} \log \left|\frac{1+\tan ^3 x}{1-\tan ^3 x}\right|\) + C

Question 11 (old).
(ii) ∫ latex]\frac{\ {cosec}^2 x}{1-\cot ^2 x}[/latex] dx
Solution:
Let I = ∫ latex]\frac{\ {cosec}^2 x}{1-\cot ^2 x}[/latex] dx
put cos x = t
⇒ – cosec² x dx = dt
∴ I = ∫ \(\frac{-d t}{1-t^2}\)
= – \(\frac{1}{2} \log \left|\frac{1+t}{1-t}\right|\)
= – \(\frac{1}{2} \log \left|\frac{1+\cot x}{1-\cot x}\right|\) + C
[∵ \(\frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|\) + C]

Question 12.
(i) ∫ \(\frac{\sin x+\cos x}{\sqrt{\sin 2 x}}\) dx
(ii) ∫ \(\frac{d x}{2 \sin ^2 x+5 \cos ^2 x}\)
Solution:
(i) Let I = ∫ \(\frac{\sin x+\cos x}{\sqrt{\sin 2 x}}\) dx
put sin x – cos x = t ……….(1)
⇒ (cos x + sin x) dx = dt
On squaring (1) ; we have
sin² x + cos² x – 2 sin x cos x = t²
⇒ 1 – sin 2x = t²
⇒ sin 2x = 1 – t²
∴ I = ∫ \(\frac{d t}{\sqrt{1-t^2}}\)
= sin^-1 (t) + C
= sin^-1 (sin x – cos x) + C

(ii) Let I = ∫ \(\frac{d x}{2 \sin ^2 x+5 \cos ^2 x}\)
Divide Numerator and denominator by cos² x ; we get
put tan x = t
⇒ sec² x dx = dt
∴ I = ∫ \(\frac{d t}{2 t^2+5}\)
= \(\frac{1}{2}\) ∫ \(\frac{d t}{t^2+\left(\sqrt{\frac{5}{2}}\right)^2}\)
= \(\frac{1}{2} \times \frac{1}{\sqrt{\frac{5}{2}}} \tan ^{-1}\left(\frac{t}{\sqrt{\frac{5}{2}}}\right)\) + C
[∵ ∫ \(\frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\) + C]
= \(\frac{1}{\sqrt{10}} \tan ^{-1}\left(\sqrt{\frac{2}{5}} \tan x\right)\) + C

Question 12 (old).
(ii) ∫ \(\frac{\sin x+\cos x}{9+16 \sin 2 x}\) dx
Solution:
Let I = ∫ \(\frac{\sin x+\cos x}{9+16 \sin 2 x}\) dx

Question 13.
(i) ∫ \(\frac{\cos x}{\cos 3 x}\) dx
(ii) ∫ \(\frac{d x}{2-3 \cos 2 x}\)
Solution:
(i) Let I = ∫ \(\frac{\cos x}{\cos 3 x}\) dx

(ii) Let I = ∫ \(\frac{d x}{2-3 \cos 2 x}\)
= ∫ \(\frac{d x}{2-3\left(1-2 \sin ^2 x\right)}\)
= ∫ \(\frac{d x}{-1+6 \sin ^2 x}\)
Divide numerator and denominator by cos² x ; we have
= ∫ \(\frac{\sec ^2 x d x}{-\sec ^2 x+6 \tan ^2 x}\)
=∫ \(\frac{\sec ^2 x d x}{-1-\tan ^2 x+6 \tan ^2 x}\)

Question 13 (old).
(ii) ∫ \(\frac{d x}{11 \sin ^2 x+4 \cos ^2 x+5}\)
Solution:
Let I = ∫ \(\frac{d x}{11 \sin ^2 x+4 \cos ^2 x+5}\)
Divide numerator and denominator by cos² x ; we have

Question 14.
(i) ∫ \(\frac{d x}{e^x-e^{-x}}\)
(ii) ∫ \(\sqrt{\frac{1+x}{1-x}}\) dx (NCERT Exemplar)
Solution:
(i) Let I = ∫ \(\frac{d x}{e^x-e^{-x}}\)
= ∫ \(\frac{d x}{e^x-\frac{1}{e^x}}\)
= ∫ \(\frac{e^x d x}{e^{2 x}-1}\)
put e^x = t
⇒ e^x dx = dt
= ∫ \(\frac{d t}{t^2-1^2}\)
= \(\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|\) + C
= \(\frac{1}{2} \log \left|\frac{e^x-1}{e^x+1}\right|\) + C

(ii) Let I = ∫ \(\sqrt{\frac{1+x}{1-x}}\) dx
put x = cos θ
⇒ dx = – sin θ dθ

= – [θ + sin θ] + C
= – cos^-1 x – \(\sqrt{1-x^{2}}\) + C
[∵ cos θ = x
⇒ θ = cos^-1 x
and sin θ = \(\sqrt{1-\cos ^2 \theta}\)
= \(\sqrt{1-x^2}\)]

Question 15.
(i) ∫ \(\frac{x-1}{\sqrt{x^2-1}}\) dx (NCERT)
(ii) ∫ \(\frac{x+2}{\sqrt{x^2-1}}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{x-1}{\sqrt{x^2-1}}\) dx
= \(\int \frac{x d x}{\sqrt{x^2-1}}-\int \frac{d x}{\sqrt{x^2-1}}\)
= I₁ – \(\frac{d x}{\sqrt{x^2-1}}\) …..(1)
where I₁ = ∫ \(\frac{x d x}{\sqrt{x^2-1}}\) ……….(1)
put x² = t
⇒ 2x dx = dt
∴ I₁ = ∫ \(\frac{d t}{2 \sqrt{t-1}}\)
= \(\frac{1}{2}\) (t – 1)^-1/2 dt
= \(\frac{1}{2} \frac{(t-1)^{1 / 2}}{1 / 2}\)
= \(\sqrt{x^{2}-1}\)
I = \(\sqrt{x^2-1}-\log \left|x+\sqrt{x^2-1}\right|\) + C

(ii) Let I = ∫ \(\frac{x+2}{\sqrt{x^2-1}}\) dx
= \(\int \frac{x d x}{\sqrt{x^2-1}}+2 \int \frac{d x}{\sqrt{x^2-1}}\)
= \(\frac{1}{2}\) ∫ (x^{2} – 1)^{\(-\frac{1}{2}\)} (2x dx) + 2 ∫ \(\frac{d x}{\sqrt{x^2-1^2}}\)
= \(\frac{1}{2} \frac{\left(x^2-1\right)^{\frac{-1}{2}+1}}{\left(\frac{-1}{2}+1\right)}\) + 2 log |x + \(\sqrt{x^2-1}\)| + C
[∵ ∫ [f(x)]ⁿ f'(x) dx = \(\frac{(f(x))^{n+1}}{n+1}\) + C]
= \(\sqrt{x^2-1}\) + 2 log |x + \(\sqrt{x^2-1}\)| + C

Question 16.
(i) ∫ \(\sqrt{\frac{a+x}{a-x}}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{\sqrt{x^2-a^2}}{x}\) dx
Solution:
(i) Let I = ∫ \(\sqrt{\frac{a+x}{a-x}}\) dx
∴ I = ∫ \(\sqrt{\frac{(a+x)}{a-x} \times \frac{a+x}{a+x}}\) dx
= ∫ \(\frac{a+x}{\sqrt{a^2-x^2}}\) dx
= ∫ \(a \int \frac{d x}{\sqrt{a^2-x^2}}+\int \frac{x d x}{\sqrt{a^2-x^2}}\)
∴ I = a sin^-1 \(\frac{x}{a}\) + I₁
where I₁ = ∫ \(\frac{d t}{2 \sqrt{a^2-t}}\)
= \(\frac{1}{2}\) ∫ (a² – t)^-1/2 dt
= – \(\frac{1}{2} \frac{\left(a^2-t\right)^{1 / 2}}{1 / 2}\)
= – \(\sqrt{a^2-x^2}\)
∴ From (1) ; we have
∴ I = a sin^-1 \(\frac{x}{a}\) – \(\sqrt{a^2-x^2}\) + C

(ii) Let I =∫ \(\frac{\sqrt{x^2-a^2}}{x}\) dx

Question 17.
If ∫ \(\frac{2^x}{\sqrt{1-4^x}}\) dx = k sin^-1 (2^x) + C, then what is the value of k?
Solution:
Let I = ∫ \(\frac{2^x}{\sqrt{1-4^x}}\) dx
= ∫ \(\frac{2^x d x}{\sqrt{1-\left(2^x\right)^2}}\)
put 2^x = t
⇒ 2^x log 2 dx = dt
⇒ I = ∫ \(\frac{d t}{\log 2 \sqrt{1-t^2}}\)
= \(\frac{1}{\log 2} \int \frac{d t}{\sqrt{1-t^2}}\)
I = \(\frac{1}{\log 2}\) sin^-1 t + C
[∵ ∫ \(\frac{d x}{\sqrt{a^2-x^2}}\) = sin^-1 \(\frac{x}{a}\) + C
∴ I = \(\frac{1}{\log 2}\) sin^-1 (2^x) + C ………….(1)
Also given
∫ \(\frac{2^x d x}{1-4 x}\) = k sin^-1 (2^x) + C ………..(2)
From (1) and (2) ; we have
k = \(\frac{1}{\log 2}\)

Parents can use ML Aggarwal Class 12 ISC Solutions Chapter 8 Integrals Ex 8.6 to provide additional support to their children.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.6

Very Short answer type questions (1 to 3):

Evaluate the following (1 to 8) integrals:

Question 1.
(i) ∫ tan (2 – 3x) dx
(ii) ∫ cot (7 – 4x) dx
Solution:
(i) Let I = ∫ \(\frac{\sin (2-3 x)}{\cos (2-3 x)}\) dx
put cos (2 – 3x) = t
⇒ – sin (2 – 3x) (- 3) dx = dt
= ∫ \(\frac{d t}{3}\)
= \(\frac{1}{3}\) log |t| + C
= \(\frac{1}{3}\) log |cos (2 – 3x)| + C

(ii) Let I = ∫ cot (7 – 4x) dx
= ∫ \(\frac{\cos (7 x-4)}{\sin (7 x-4)}\) dx
put sin (7x – 4) = t
⇒ cos (7x – 4) . 7 dx = dt
= ∫ \(\frac{d t}{7 t}\)
= \(\frac{1}{7}\) log |t| + C
= \(\frac{1}{7}\) log |sin (7x – 4)| + C

Question 2.
(i) ∫ \(\frac{1+\tan ^2 x}{2 \tan x}\) dx
(ii)∫ \(\frac{1}{\sin x \cos x}\) dx
Solution:
(i) Let I = ∫ \(\frac{1+\tan ^2 x}{2 \tan x}\) dx
= \(\frac{1}{2} \int \frac{\sec ^2 x d x}{\tan x}\)
= \(\frac{1}{2}\) log |tan x| + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]

Aliter:
Let I = ∫ \(\frac{\sec ^2 x d x}{2 \tan x}\)
= ∫ \(\frac{d x}{2 \sin x \cos x}\)
= ∫ \(\frac{d x}{\sin 2 x}\)
= ∫ cosec 2x dx
= \(\frac{\log |\ {cosec} 2 x-\cot 2 x|}{2}\) + C

(ii) Let I = ∫ \(\frac{d x}{\sin x \cos x}\)
= ∫ \(\frac{2 d x}{\sin 2 x}\)
= 2 ∫ cosec 2x dx
= \(\frac{2 \log |\ {cosec} 2 x-\cot 2 x|}{2}\) + C
= log |cosec 2x – cot 2x| + C

Question 3.
(i) ∫ \(\frac{\cos 2 x}{\sin x}\) dx
(ii) ∫ \(\frac{\sin 2 x}{\sin 4 x}\) dx
(iii) ∫ \(\frac{\sin ^2 x-\cos ^2 x}{\sin x \cos x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\cos 2 x}{\sin x}\) dx
= ∫ \(\frac{\left(1-2 \sin ^2 x\right)}{\sin x}\) dx
= ∫ cosec x dx – 2 ∫ sin x dx + C
= log |cosec x – cot x| + 2 cos x + C

(ii) Let I = ∫ \(\frac{\sin 2 x}{\sin 4 x}\) dx
= ∫ \(\frac{\sin 2 x d x}{2 \sin 2 x \cos 2 x}\) dx
= \(\frac{1}{2}\) ∫ sec 2x dx
= \(\frac{1}{2} \frac{\log |\sec 2 x+\tan 2 x|}{2}\) + C
= \(\frac{1}{4}\) log |sec 2x + tan 2x| + C

(iii) Let I = ∫ \(\frac{\sin ^2 x-\cos ^2 x}{\sin x \cos x}\) dx
= \(\int \frac{\sin x}{\cos x} d x-\int \frac{\cos x}{\sin x} d x\)
= – \(\int \frac{-\sin x}{\cos x} d x-\int \frac{\cos x d x}{\sin x}\)
= – log |cos x| – log |sin x| + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]
= log |sec x| – log |sin x| + C
[∵ a log b = log b^q]

Question 4.
(i) ∫ \(\frac{\sin x}{\sin (x-\alpha)}\) dx
(ii) ∫ \(\frac{\cos x}{\cos (x+\alpha)}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin x}{\sin (x-\alpha)}\) dx
= ∫ \(\frac{\sin (x-\alpha+\alpha) d x}{\sin (x-\alpha)}\)
= ∫ cos α dx + sin α ∫ \(\frac{\cos (x-\alpha)}{\sin (x-\alpha)}\) dx
= x cos α + sin α log |sin (x – α)| + C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]

(ii) Let I = ∫ \(\frac{\cos x}{\cos (x+\alpha)}\) dx
= ∫ \(\frac{\cos x}{\cos (x+\alpha)}\) dx
= ∫ \(\frac{\cos (x+\alpha-\alpha)}{\cos (x+\alpha)}\) dx
= ∫ \(\left[\frac{\cos (x+\alpha) \cos \alpha-\sin (x+\alpha) \sin \alpha}{\cos (x+\alpha)}\right]\) dx
= ∫ cos α dx – sin α ∫ \(\frac{\sin (x+\alpha) d x}{\cos (x+\alpha)}\)
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]
= x cos α + sin α log |cos (x + α)| + C

Question 5.
(i) ∫ \(\frac{d x}{\sin (x-\alpha) \sin (x-\beta)}\)
(ii) ∫ \(\frac{d x}{\cos (x-a) \cos (x-b)}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sin (x-\alpha) \sin (x-\beta)}\)

(ii) Let I = ∫ \(\frac{d x}{\cos (x-a) \cos (x-b)}\)

Question 6.
(i) ∫ \(\frac{d x}{\sin (x-a) \cos (x-b)}\)
(ii) ∫ tan³ x dx (ISC 2016)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sin (x-a) \cos (x-b)}\)

(ii) Let I = ∫ tan³ x dx
= ∫ tan x (sec² – 1) dx
= ∫ tan x (sec² x dx) – ∫ tan x dx
= \(\frac{\tan ^2 x}{2}\) + log |cos x| + C
[∵ ∫ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\), n ≠ 1]

Question 7.
(i) ∫ \(\frac{\sin 2 x}{\sin 5 x \sin 3 x}\) dx
(ii) ∫ \(\frac{1}{\sin x \cos ^2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin 2 x}{\sin 5 x \sin 3 x}\) dx

(ii) Let I = ∫ \(\frac{1}{\sin x \cos ^2 x}\) dx
= ∫ \(\frac{\left(\sin ^2 x+\cos ^2 x\right) d x}{\sin x \cos ^2 x}\)
= ∫ tan x sec x dx + ∫ cosec x dx
= sec x + log |cosec x – cot x| + C

Question 8.
(i) ∫ \(\frac{d x}{\sqrt{1-\sin x}}\)
(ii) ∫ \(\frac{\sin x+\cos x}{1+\sin 2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{1-\sin x}}\)
= ∫ \(\frac{d x}{\sqrt{1-\cos \left(\frac{\pi}{2}-x\right)}}\)
= ∫ \(\frac{d x}{\sqrt{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}}\)
= \(\frac{1}{\sqrt{2}} \int \frac{d x}{\sin \left(\frac{\pi}{4}-\frac{x}{2}\right)}\)
= \(\frac{1}{\sqrt{2}} \int \ {cosec}\left(\frac{\pi}{4}-\frac{x}{2}\right)\)
= \(\frac{1}{\sqrt{2}} \frac{\log \left|\ {cosec}\left(\frac{\pi}{4}-\frac{x}{2}\right)-\cot \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|}{-\frac{1}{2}}\) + C
= – √2 log \(\left|\ {cosec}\left(\frac{\pi}{4}-\frac{x}{2}\right)-\cot \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|\) + C

Aliter:

(ii) Let I = ∫ \(\frac{\sin x+\cos x}{1+\sin 2 x}\) dx

Question 9.
∫ \(\frac{\sqrt{2} \sin x}{\sin \left(x-\frac{\pi}{4}\right)}\) dx = ax + b log |sin (x – \(\frac{\pi}{4}\))| + C, find the values of a and b.
Solution:
Let I = ∫ \(\frac{\sqrt{2} \sin x}{\sin \left(x-\frac{\pi}{4}\right)}\) dx

Also given I = ∫ \(\frac{\sqrt{2} \sin x}{\sin \left(x-\frac{\pi}{4}\right)}\) dx
= ax + b log |sin (x – \(\frac{\pi}{4}\))| + C …………(2)
∴ From (1) and (2) ; we have
a = 1 ;
b = 1.

The availability of ML Aggarwal Class 12 Solutions ISC Chapter 8 Integrals Ex 8.5 encourages students to tackle difficult exercises.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.5

Very Short answer type questions (1 to 3):

Evaluate the following (1 to 23) integrals:

Question 1.
(i) ∫ 2x sin (x² + 1) dx
(ii) ∫ x³ cos (x⁴) dt
Solution:
(i) Let I = ∫ 2x sin (x² + 1) dx
put x² + 1 = t
⇒ 2x dx = dt
∴ I = ∫ sin t dt
= – cos t + C
= – cos (x² + 1) + C

(ii) Let I = ∫ x³ cos (x⁴) dt
put x⁴ = t
⇒ d (x⁴) = dt
⇒ 4x³ dx = dt
⇒ x³ dx = \(\frac{1}{4}\) dt
∴ I = ∫ cos x⁴ (x³ dx)
= ∫ cos t . \(\frac{d t}{4}\)
= \(\frac{1}{4}\) sin t + c
= \(\frac{1}{4}\) sin x⁴ + c

Question 2.
(i) ∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx
(ii) ∫ \(\frac{\sin \sqrt{x}}{\sqrt{x}}\) dx
(iii) ∫ \(\frac{\sec ^2 \sqrt{x}}{\sqrt{x}}\) dx
(iv) ∫ \(\frac{\ {cosec}^2 \sqrt{x}}{\sqrt{x}}\) dx
Solution:
(i) Let I = ∫ \(\frac{\cos \sqrt{x}}{\sqrt{x}}\) dx
put √x = t
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ \(\frac{d x}{\sqrt{x}}\) = 2 dt
∴ I = ∫ cos t (2 dt) = 2 sin t + C

(ii) Let I = ∫ \(\frac{\sin \sqrt{x}}{\sqrt{x}}\) dx
put √x = t
⇒ \(\frac{1}{\sqrt{x}}\) dx = dt
∴ I = ∫ sin t dt
= – cos t + c
= – cos √x + c

(iii) Let I = ∫ \(\frac{\sec ^2 \sqrt{x}}{\sqrt{x}}\) dx
put √x = t
⇒ d (√x) = dt
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
⇒ \(\frac{d x}{\sqrt{x}}\) = 2 dt
∴ I = 2 ∫ sec² t dt
= 2 tan t + c
= 2 tan √x + c

(iv) Let I = ∫ \(\frac{\ {cosec}^2 \sqrt{x}}{\sqrt{x}}\) dx
put √x = t
⇒ \(\frac{1}{2 \sqrt{x}}\) dx = dt
∴ I = ∫ cosec² t (2 dt) = – 2 cot t + C
= – 2 cot √x + C.

Question 3.
(i) ∫ sin x . e^{cos x} dx
(ii) ∫ \(\frac{e^{2 x}}{1+e^x}\) dx
Solution:
(i) Let I = ∫ sin x . e^{cos x} dx
put cos x = t
⇒ – sin x dx = dt
= ∫ e^t (- dt)
= – e^t + C
= – e^{cos x} + C

(ii) Let I = ∫ \(\frac{e^{2 x}}{1+e^x}\) dx
= ∫ \(\frac{e^x \cdot e^x d x}{1+e^x}\) dx
put e^x = t
⇒ d (e^x) = dt
⇒ e^x dx = dt
= ∫ \(\frac{t d t}{1+t}\)
= ∫ \(\frac{1+t-1}{1+t}\)
= ∫ \(\left[1-\frac{1}{t+1}\right]\) dt
= t – log |1 + t| + c
= e^x – log |1 + e^x| + c

Question 4.
(i) ∫ sin x sin (cos x) dx (NCERT)
(ii) ∫ x³ sec² (x⁴ + 3) dx
Solution:
(i) Let I = ∫ sin x sin (cos x) dx
put cos x = t
⇒ – sin x dx = dt
∴ I = ∫ sin t (- dt)
= cos t + C
= cos (cos x) + C

(ii) Let I = ∫ x³ sec² (x⁴ + 3) dx
put x⁴ + 3 = t
⇒ 4x³ dx = dt
= ∫ sec² t (\(\frac{d t}{4}\)) = \(\frac{1}{4}\) tan t + C
= \(\frac{1}{4}\) tan (x⁴ + 4) + C

Question 5.
(i) ∫ \(\frac{1}{x^2} \tan ^2\left(\frac{1}{x}\right)\) dx
(ii) ∫ \(\frac{1}{x^2} \sin ^2\left(\frac{1}{x}\right)\) dx (ISC 2017)
(iii) ∫ 2x sec³ (x² + 5) tan (x² + 5) dx
Solution:
(i) Let I = ∫ \(\frac{1}{x^2} \tan ^2\left(\frac{1}{x}\right)\) dx
put \(\frac{1}{x}\) = t
⇒ – \(\frac{1}{x^2}\) dx = dt
∴ I = ∫ tan² t (- dt)
= ∫ – (sec² t dt + ∫ dt + C
= – tan t + t + C
= – tan (\(\frac{1}{x}\)) + \(\frac{1}{x}\) + C

(ii) Let I = ∫ \(\frac{1}{x^2} \sin ^2\left(\frac{1}{x}\right)\) dx

(iii) Let I = ∫ 2x sec³ (x² + 5) tan (x² + 5) dx
put x² + 5 = t
⇒ 2x dx = dt
= ∫ sec³ t tan t dt
= ∫ sec² t (sec t tan t dt)
= \(\frac{\sec ^3 t}{3}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{1}{3}\) sec³ (x² + 5) + C

Question 6.
(i) ∫ \(\frac{\sin (\log x)}{x}\) dx
(ii) ∫ \(\frac{\ {cosec}^2(\log x)}{x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin (\log x)}{x}\) dx
put log x = t
⇒ d (log x) dt
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫ sin t dt
= – cos t + c
= – cos (log x) + c

(ii) Let I = ∫ \(\frac{\ {cosec}^2(\log x)}{x}\) dx
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫ cosec² t dt
= – cot t + C
= – cot (log x) + C

Question 7.
(i) ∫ \(\frac{e^{\tan ^{-1} x}}{1+x^2}\) dx
(ii) ∫ \(\frac{\sin \left(2 \tan ^{-1} x\right)}{1+x^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{e^{\tan ^{-1} x}}{1+x^2}\) dx
put tan^-1 x = t
⇒ \(\frac{1}{1+x^2}\) dx = dt
= ∫ e^t dt
= e^t + C
= e^{tan-1 x} + C

(ii) Let I = ∫ \(\frac{\sin \left(2 \tan ^{-1} x\right)}{1+x^2}\) dx
put tan^-1 x = t
⇒ \(\frac{1}{1+x^2}\) dx = dt
∴ I = ∫ sin 2t dt
= – \(\frac{\cos 2 t}{2}\) + C
= – \(\frac{1}{2}\) cos (2 tan^-1 x) + C

Question 8.
(i) ∫ \(\frac{e^{-1 / x}+1}{x^2}\) dx
(ii) ∫ \(\frac{e^{m \tan ^{-1} x}}{1+x^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{e^{-1 / x}+1}{x^2}\) dx
= ∫ (e^-1/x + 1) \(\frac{1}{x^{2}}\) dx
put – \(\frac{1}{x}\) = t
⇒ \(\frac{1}{x^{2}}\) dx = dt
= ∫ (e^t + 1) dt
= e^t + t + C
= e^{– 1/x} – \(\frac{1}{x}\) + C

(ii) Let I = ∫ \(\frac{e^{m \tan ^{-1} x}}{1+x^2}\) dx
put tan^-1 x = t
⇒ \(\frac{1}{1+x^2}\) dx = dt
∴ I = ∫ e^mt dt
= \(\frac{e^{m t}}{m}\) + C
= \(\frac{e^{m \tan ^{-1} x}}{m}\) + C

Question 9.
(i) ∫ e^x cosec² (e^x) dt
(ii) ∫ \(\frac{(x+1) e^x}{\cot ^2\left(x e^x\right)}\) dx
Solution:
(i) Let I = ∫ e^x cosec² (e^x) dt
put e^x = t
⇒ e^x dx = dt
∴ I = ∫ cosec² t dt
= – cot t + C
= – cot (e^x) + C

(ii) Let I = ∫ \(\frac{(x+1) e^x}{\cot ^2\left(x e^x\right)}\) dx
put x e^x = t
d (x e^x) = dt
⇒ (x e^x + e^x) dx = dt
⇒ (x + 1) e^x dx = dt
∴ I = ∫ \(\int \frac{d t}{\cot ^2 t}\)
= ∫ tan² t dt
= ∫ (sec² t – 1) dt
= tan t – t + C
= tan (x e^x) – x e^x + C

Question 10.
(i) ∫ x² e^x3 (sin e^x3) dx
(ii) ∫ \(\frac{\sin ^2(\log x)}{x}\) dx
(iii) ∫ \(\frac{\sqrt{\tan x}}{\sin ^2 x}\) dx
Solution:
(i) Let I = ∫ x² e^x3 (sin e^x3) dx
put e^x3 = t
⇒ e^x3 . 3x² dx = dt
∴ I = \(\frac{1}{3}\) ∫ sin t dt
= – \(\frac{\cos t}{3}\) + C
= – \(\frac{1}{3}\) cos (e^x3) + C

(ii) Let I = ∫ \(\frac{\sin ^2(\log x)}{x}\) dx
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫ sin² t dt
= ∫ \(\frac{1-\cos 2 t}{2}\) dt
= \(\frac{1}{2}\left[t-\frac{\sin 2 t}{2}\right]\) + C
= \(\frac{1}{2}\) [log x – \(\frac{1}{2}\) sin (2 log x)] + C

(iii) Let I = ∫ \(\frac{\sqrt{\tan x}}{\sin ^2 x}\) dx
= ∫ \(\frac{\sqrt{\tan x}}{\frac{\sin ^2 x}{\cos ^2 x} \cos ^2 x}\) dx
= ∫ \(\frac{\sqrt{\tan x}}{\tan ^2 x}\) sec² x dx
= ∫ (tan x)^-3/2 sec² x dx
put tan x = t
⇒ sec² x dx = dt
= ∫ t^-3/2 dt
= \(\frac{t^{-3 / 2}+1}{-\frac{3}{2}+1}\) + C
= – 2 t^-1/2 + C
= – \(\frac{2}{\sqrt{\tan x}}\) + C

Question 11.
(i) ∫ x sin³ (x²) cos (x²) dx
(ii) ∫ tan 2x sec 2x dx (NCERT)
Solution:
(i) Let I = ∫ x sin³ (x²) cos (x²) dx
put sin x² = t
⇒ cos (x²) . 2x dx = dt
= ∫ t³ \(\frac{d t}{2}\)
= \(\frac{1}{2}\) \(\frac{t^4}{4}\) + C
= \(\frac{t^4}{8}\) + C
= \(\frac{1}{8}\) sin⁴ (x² + C

(ii) Let I = ∫ tan³ 2x sec 2x dx
= ∫ tan² 2x (tan 2x sec 2x) dx
= ∫ (sec² 2x – 1) tan 2x sec 2x dx
put sec 2x = t
⇒ (sec 2x tan 2x) 2 dx = dt
= ∫ (t² – 1) \(\frac{d t}{2}\)
= \(\frac{1}{2}\left[\frac{t^3}{3}-t\right]\) + C
= \(\frac{t^3}{6}-\frac{1}{2} t\) + C
= \(\frac{1}{6}\) sec³ 2x – \(\frac{1}{2}\) sec 2x + C

Question 12.
(i) ∫ sin³ (2x + 1) dx (NCERT)
(ii) ∫ sin³ x cos² x dx (NCERT)
Solution:
(i) Let I = ∫ sin³ (2x + 1) dx
= ∫ sin² (2x + 1) sin (2x + 1) dx
= ∫ (1 – cos² (2x + 1)) sin (2x + 1) dx
put cos (2x + 1) = t
⇒ – 2 sin (2x + 1) dx = dt
∴ I = ∫ (1 – t²) \(\frac{d t}{- 2}\)
= \(\) + C
= – \(\frac{1}{2}\) cos (2x + 1) + \(\frac{1}{6}\) cos³(2x + 1) + C

(ii) Let I = ∫ sin³ x cos² x dx
= ∫ \(\frac{\cos ^3 x}{\sin ^2 x}\) dx
= ∫ \(\frac{\left(1-\sin ^2 x\right) \cos x}{\sin ^2 x}\) dx
put sin x = t
⇒ cos x dx = dt
∴ I = ∫ \(\frac{\left(1-t^2\right) d t}{t^2}\)
= ∫ \(\frac{1}{t^2}\) dt – ∫ dt
= – \(\frac{1}{t}\) – t + C
= – \(\frac{1}{sin x}\) – sin x + C
= – cosec x – sin x + C

Question 14.
(i) ∫ \(\frac{\cos ^3 x}{\sqrt{\sin x}}\) dx
(ii) ∫ \(\frac{\sin x \cos ^3 x}{1+\cos ^2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\cos ^3 x}{\sqrt{\sin x}}\) dx
= ∫ \(\frac{\cos ^2 x \cos x d x}{\sqrt{\sin x}}\)
= ∫ \(\frac{\left(1-\sin ^2 x\right) \cos x d x}{\sqrt{\sin x}}\)
put sin x = t
⇒ cos x dx = dt
= ∫ \(\frac{1-t^2}{\sqrt{t}}\) dt
= ∫ \(\left[\frac{1}{\sqrt{t}}-t^{3 / 2}\right]\) dt
put sin x = t
⇒ cos x dx = dt
= ∫ \(\frac{1-t^2}{\sqrt{t}}\) dt
= ∫ \(\left[\frac{1}{\sqrt{t}}-t^{3 / 2}\right]\) dt
= \(\frac{t^{\frac{-1}{2}+1}}{\left(\frac{-1}{2}+1\right)}-\frac{t^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}\) + c
= 2√t – \(\frac{2}{5}\) t^5/2 + c
= 2 \(\sqrt{sin x}\) – \(\frac{2}{5}\) t^5/2 + c

(ii) Let I = ∫ \(\frac{\sin x \cos ^3 x}{1+\cos ^2 x}\) dx
put cos x = t
⇒ – sin x dx = dt

Question 15.
(i) ∫ cosec x log (cosec x – cot x) dx
(ii) ∫ \(\frac{\log \left(\tan \frac{x}{2}\right)}{\sin x}\) dx
Solution:
(i) Let I = ∫ cosec x log (cosec x – cot x) dx
put log (cosec x – cot x) = t
⇒ d {log (cosec x – cot x)} = dt
[- cot x cosec x + cosec² x] dx = dt
⇒ cosec x dx = \(\frac{1}{\ {cosec} x-\cot x}\) dt
∴ I = ∫ t dt
= \(\frac{t^{2}}{2}\) + c
= \(\frac{1}{2}\) [log (cosec x – cot x)]² + c

(ii) Let I = ∫ \(\frac{\log \left(\tan \frac{x}{2}\right)}{\sin x}\) dx
put log (tan \(\frac{x}{2}\)) = t
⇒ \(\frac{1}{\tan \frac{x}{2}} \sec ^2 \frac{x}{2} \cdot \frac{1}{2}\) dx = dt
⇒ \(\frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\) dx = dt
⇒ \(\frac{1}{\sin x}\) dx = dt
∴ I = ∫ t dt
= \(\frac{t^{2}}{2}\) + c
= \(\frac{\left[\log \left(\tan \frac{x}{2}\right)\right]^2}{2}\) + C

Question 16.
(i) ∫ sec⁴ x dx
(ii) ∫ tan² x sec⁴ x dx (NCERT Exemplar)
Solution:
(i) Let I = ∫ sec⁴ x dx
= ∫ sec² x (1 + tan² x) dx
put tan x = t
⇒ sec² x dx = dt
∴ I = ∫ (1 + t² dt
= t + \(\frac{t^{3}}{3}\) + C
= tan x + \(\frac{\tan ^3 x}{3}\) + C

(ii) Let I = ∫ tan² x sec⁴ x dx
= ∫ tan² x . sec² x . sec² x dx
= ∫ tan² x (1 + tan² x) . sec² x dx
put tan x = t
sec² x dx = dt
= ∫ t² (1 + t²) dt
= \(\frac{t^3}{3}+\frac{t^5}{5}\) + C
= \(\frac{1}{3}\) tan³ x + \(\frac{1}{5}\) tan⁵ x + C

Question 17.
(i) ∫ \(\frac{\tan ^5 \sqrt{x} \sec ^2 \sqrt{x}}{\sqrt{x}}\) dx
(ii) ∫ \(\frac{\cos x-\sin x}{\sqrt{1+\sin 2 x}}\) dx
Solution:
(i) Let I = ∫ \(\frac{\tan ^5 \sqrt{x} \sec ^2 \sqrt{x}}{\sqrt{x}}\) dx
put tan √x = t
sec² √x \(\left(\frac{1}{2 \sqrt{x}}\right)\) dx = dt
∴ I = ∫ t⁵ (2 dt)
= \(\frac{2 t^6}{6}\) + C
= \(\frac{1}{3}\) tan⁶ (√x) + C

(ii) Let I = ∫ \(\frac{\cos x-\sin x}{\sqrt{1+\sin 2 x}}\) dx
= ∫ \(\frac{(\cos x-\sin x) d x}{\sqrt{\cos ^2 x+\sin ^2 x+2 \sin x \cos x}}\)
= ∫ \(\frac{(\cos x-\sin x) d x}{\sqrt{(\cos x+\sin x)^2}}\)
put cos x + sin x = t
⇒ (- sin x + cos x) dx = dt
= ∫ \(\frac{d t}{t}\)
= log |t| + C
= log |cos x + sin x| + C

Question 18.
(i) ∫ \(\frac{x}{\sqrt{x+4}}\) dx (NCERT)
(ii) ∫ x \(\sqrt{x+2}\) dx (NCERT)
Solution:
(i) Let I = ∫ \(\frac{x d x}{\sqrt{x+4}}\)
= ∫ \(\frac{(x+4-4)}{\sqrt{x+4}}\) dx
= ∫ \(\sqrt{x+4}\) dx – 4 ∫ (x + 4)^{\(-\frac{1}{2}\)} dx
= \(\frac{2}{3}\) (x + 4)\(\frac{3}{2}\) – 4 \(\frac{(x+4)^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)}\) + C
= \(\frac{2}{3}\) (x + 4)\(\frac{3}{2}\) – 8 \(\sqrt{x+4}\) + C
∴ I = \(\frac{2}{3} \sqrt{x+4}\) (x + 4 – 12)
= \(\frac{2}{3}\) (x – 8) \(\sqrt{x+4}\) + C

(ii) Let I = ∫ x \(\sqrt{x+2}\) dx
= ∫ (x + 2 – 2) \(\sqrt{x+2}\) dx
= ∫ (x + 2)^{\(\frac{3}{2}\)} dx – 2 ∫ (x + 2)^{\(\frac{1}{2}\)} dx
= \(\frac{2}{5}(x+2)^{\frac{5}{2}}-2 \times \frac{2}{3}(x+2)^{\frac{3}{2}}\) + C
= \(\frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}}\) + C

Question 19.
(i) ∫ \(\frac{x}{\left(x^2+1\right)^2}\) dx
(ii) ∫ \(\frac{x^2}{(4+x)^{3 / 2}}\) dx
Solution:
(i) Let I = ∫ \(\frac{x}{\left(x^2+1\right)^2}\) dx
put x² = t
⇒ 2x dx = dt
∴ I = ∫ \(\frac{d t}{2(t+1)^2}\)
= \(\frac{1}{2} \frac{(t+1)^{-2+1}}{(-2+1)}\) + C
= – \(\frac{1}{2(t+1)}\) + C
= – \(\frac{1}{2\left(x^2+1\right)}\) + C

(ii) Let I = ∫ \(\frac{x^2}{(4+x)^{3 / 2}}\) dx
put 4 + x = t
⇒ dx = dt

Question 20.
(i) ∫ 4x³ \(\sqrt{5-x^2}\) dx
(ii) ∫ \(\frac{d x}{x-\sqrt{x}}\) (NCERT)
Solution:
(i) Let I = ∫ 4x³ \(\sqrt{5-x^2}\) dx
put 5 – x² = t
⇒ x² = 5 – t
⇒ 2x dx = – dt
∴ I = ∫ 2 (5 – t) √t (- dt)
= – 10 t^{\(\frac{1}{2}\) + 1} / (\(\frac{1}{2}\) + 1) + 2 t^{\(\frac{5}{2}\)} / \(\frac{5}{2}\) + C
= – \(\frac{20}{3}\) t^3/2 + \(\frac{4}{5}\) t^5/2 + C
= \(\frac{4}{5}\) (5 – x²)^5/2 – \(\frac{20}{3}\) (5 – x²)^5/2 + C

(ii) Let I = ∫ \(\frac{d x}{x-\sqrt{x}}\)
put √x = t
⇒ x = t²
⇒ dx = 2t dt
∴ I = ∫ \(\frac{2 t d t}{t^2-t}\)
= ∫ \(\frac{2 t d t}{t(t-1)}\)
= 2 ∫ \(\frac{d t}{t-1}\)
= 2 log |t – 1| + C
= 2 log |√x – 1| + C

Question 21.
(i) ∫ \(\frac{x}{1+\sqrt{x}}\) dx (NCERT Exemplar)
(ii) ∫ \(\frac{x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}}\) dx
Solution:
(i) Let I = ∫ \(\frac{x}{1+\sqrt{x}}\) dx
put √x = t
⇒ x = t²
⇒ dx = 2t dt

(ii) Let I = ∫ \(\frac{x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}}\) dx
put x^1/4 = t
⇒ x = t⁴
⇒ dx = 4t³

Question 22.
(i) ∫ \(\frac{d x}{\sqrt{x+1}+\sqrt[3]{x+1}}\)
(ii) ∫ \(\frac{d x}{\sqrt{1+\sqrt{x}}}\)
Solution:
(i) Let I = ∫ \(\frac{d x}{\sqrt{x+1}+\sqrt[3]{x+1}}\)

(ii) Let I = ∫ \(\frac{d x}{\sqrt{1+\sqrt{x}}}\)
put \(\sqrt{1+\sqrt{x}}\) = t
⇒ 1 + √x = t
⇒ √x = t² – 1
⇒ x = (t² – 1)²
⇒ dx = 4t (t² – 1)
∴ I = ∫ \(\frac{4 t\left(t^2-1\right)}{t}\)
= 4 \(\left[\frac{t^3}{3}-t\right]\) + C
= \(\frac{4}{3}\) (1 + √x)^3/2 – 4 \(\sqrt{1+\sqrt{x}}\) + C

Question 23.
(i) ∫ \(\frac{\sin 2 x}{\left(a^2+b^2 \sin ^2 x\right)^2}\) dx
(ii) ∫ \(\frac{\sqrt{1+x^2}}{x^4}\) dx (NCERT Exemplar)
(iii) ∫ \(\frac{1}{x^2 \sqrt{1+x^2}}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin 2 x}{\left(a^2+b^2 \sin ^2 x\right)^2}\) dx
put a² + b² sin² x = t
⇒ 2b² sin x cos x dx = dt
⇒ b² sin 2x dx = dt

(ii) Let I = ∫ \(\frac{\sqrt{1+x^2}}{x^4}\) dx
put x = tan θ
dx = sec² θ
∴ I = ∫ \(\frac{\sqrt{1+\tan ^2 \theta}}{\tan ^4 \theta}\) sec² θ dθ
= ∫ \(\frac{\sec ^3 \theta d \theta}{\tan ^4 \theta}\)
= ∫ \(\frac{\frac{1}{\cos ^3 \theta}}{\frac{\sin ^4 \theta}{\cos ^4 \theta}}\) dθ
= ∫ \(\frac{\cos \theta d \theta}{\sin ^4 \theta}\)
= ∫ (sin θ)^-4 cos θ dθ
= \(\frac{(\sin \theta)^{-4+1}}{(-4+1)}\) + C
[∵ ∫ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C]

(iii) Let I = ∫ \(\frac{1}{x^2 \sqrt{1+x^2}}\) dx

Question 24.
(i) If ∫ x e^kx2 dx = \(\frac{1}{4}\) e^2x2 + C, then find the value of k.
(ii) If ∫ x⁶ sin (5x)⁷ dx = \(\frac{k}{5}\) cos (5x⁷) + C, then what is the value of k ?
Solution:
(i) Let I = ∫ x e^kx2 dx
put x² = t
2x dx = dt
= ∫ e^kt \(\frac{d t}{2}\)
= \(\frac{1}{2} \frac{e^{k t}}{k}\) + C
= \(\frac{1}{2 k}\) e^kx2 + C ……….(1)
Also given
I = ∫ x e^kx2 dx
= \(\frac{1}{4}\) e^2x2 + C …………..(2)
From (1) and (2) ; we have
\(\)
⇒ 2k = 4
⇒ k = 2

(ii) put x⁷ = t
⇒ 7x⁶ dx = dt
⇒ x⁶ dx = \(\frac{d t}{7}\)
∴ ∫ x⁶ sin (5x⁷) dx = ∫ sin 5t \(\frac{d t}{7}\)
= – \(\frac{\cos 5 x^7}{35}\) + C ……….(1)
Also given,
∫ x⁶ sin (5x⁷) dx = \(\frac{k}{5}\) cos (5x⁷) + C ………..(2)
∴ from (1) and (2) ; we have
\(\frac{k}{5}=-\frac{1}{35}\)
k = – \(\frac{1}{7}\)

Regular engagement with ML Aggarwal Class 12 Solutions Chapter 8 Integrals Ex 8.4 can boost students confidence in the subject.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.4

Very Short answer type questions (1 to 9):

Evaluate the following (1 to 21) integrals:

Question 1.
(i) ∫ x³ (1 + x⁴)³ dx
(ii) ∫ (2x + 4) \(\sqrt{x²+4x+3}\) dx
Solution:
(i) Let I = ∫ x³ (1 + x⁴)³ dx
= \(\frac{1}{4}\) ∫ (1 + x⁴)x³ dx
= \(\frac{1}{4} \frac{\left(1+x^4\right)^4}{4}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{\left(1+x^4\right)^4}{16}\) + C

(ii) Let I = ∫ (2x + 4) \(\sqrt{x²+4x+3}\) dx
= ∫ (x² + 4x + 3)^1/2 (2x + 4) dx
= \(\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{2}{3}\) (x² + 4x + 3)^3/2 + C

Question 2.
(i) ∫ x (x² + 3)^3/2 dx
(ii) ∫ \(\frac{4 x+1}{\sqrt{2 x^2+x-7}}\) dx
Solution:
(i) ∫ x (x² + 3)^3/2 dx
= \(\frac{1}{2}\) ∫ (x² + 3)^3/2 dx
= \(\frac{1}{2} \frac{\left(x^2+3\right)^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}\) + C
= \(\frac{1}{5}\) (x² + 3)^5/2 + C

(ii) Let I = ∫ \(\frac{4 x+1}{\sqrt{2 x^2+x-7}}\) dx
= ∫ (2x² + x – 7)^{\(-\frac{1}{2}\)} (4x + 1) dx
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{\left(2 x^2+x-7\right)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= 2 \(\sqrt{2 x^2+x-7}\) + C

Question 3.
(i) ∫ \(\frac{3 x+1}{\left(3 x^2+2 x-5\right)^3}\) dx
(ii) ∫ \(\frac{2 x}{\sqrt[3]{x^2+1}}\) dx
Solution:
(i) Let I = ∫ \(\frac{3 x+1}{\left(3 x^2+2 x-5\right)^3}\) dx
= ∫ (3x² + 2x – 5)^{– 3} (3x + 1) dx
= \(\frac{1}{2}\) ∫ (3x² + 2x – 5)^{– 3} (6x + 2) dx
= \(\frac{1}{2} \frac{\left(3 x^2+2 x-5\right)^{-3+1}}{(-3+1)}\) + C
= – \(\frac{1}{4}\) ∫ (3x² + 2x – 5)^{– 2} + C
= – \(\frac{1}{4\left(3 x^2+2 x-5\right)^2}\) + C

(ii) Let I = ∫ \(\frac{2 x}{\sqrt[3]{x^2+1}}\) dx
put x² + 1 = t
⇒ 2x + dx = dt
= ∫ \(\frac{d t}{t^{1 / 3}}\)
= ∫ t^{\(-\frac{1}{3}\)} dt
= \(\frac{t^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}\) + C
= \(\frac{3}{2}\) t^2/3 + C
= \(\frac{3}{2}\) (x² + 1)^2/3 + C

Question 3 (old).
(ii) ∫ (4x + 2) \(\sqrt{x^2+x-3}\) dx
Solution:
Let I = ∫ (4x + 2) \(\sqrt{x^2+x-3}\) dx
= ∫ (x² + x – 3)^{\(\frac{1}{2}\)} 2 (2x + 1) dx
= \(\frac{2\left(x^2+x-3\right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{2\left(x^2+x-3\right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
= \(\frac{4}{3}\) (x² + x – 3)^3/2 + C

Question 4.
(i) ∫ e^x (a + be^x)⁶ dx
(ii) ∫ \(\frac{(\log x)^2}{x}\) dx
Solution:
(i) Let I = ∫ e^x (a + be^x)⁶ dx
= \(\frac{1}{b}\) ∫ (a + be^x)⁶ be^x dx
= \(\frac{1}{b} \frac{\left(a+b e^x\right)^{6+1}}{6+1}\) + C
= \(\frac{1}{7 b}\) (a + be^x)⁷ + C

(ii) Let I = ∫ \(\frac{1}{x}\) (log x)² dx
put log x = t
⇒ d (log x) = dt
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫ t² dt
= \(\frac{t^{3}}{3}\) + c
= \(\frac{(\log x)^3}{3}\) + c

Question 5.
(i) ∫ \(\frac{\sqrt{3+\log x}}{x}\) dx
(ii) ∫ \(\frac{3(x+1)(x+\log x)^2}{x}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sqrt{3+\log x}}{x}\) dx
= ∫ (3 + log x)^1/2 \(\frac{1}{x}\) dx
[∵ \(\frac{d}{d x}\) (3 + log x) = 0 + \(\frac{1}{x}\) = \(\frac{1}{x}\)]
= \(\frac{(3+\log x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{2}{3}\) (3 + log x)^3/2 + C

(ii) Let I = ∫ \(\frac{3(x+1)(x+\log x)^2}{x}\) dx
= 3 ∫ (x + log x)² (1 + \(\frac{1}{x}\)) dx
[∵ \(\frac{d}{d x}\) (x + log x) = 1 + \(\frac{1}{x}\)]
= 3 \(\frac{(x+\log x)^3}{3}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= (x + log x)³ + C

Question 6.
(i) ∫ \(\frac{\sqrt{\tan ^{-1} x}}{1+x^2}\) dx
(ii) ∫ \(\frac{\left(\sin ^{-1} 2 x\right)^3}{\sqrt{1-4 x^2}}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sqrt{\tan ^{-1} x}}{1+x^2}\) dx
= ∫ \(\left(\tan ^{-1} x\right)^{\frac{1}{2}} \cdot \frac{1}{1+x^2}\) dx
= \(\frac{\left(\tan ^{-1} x\right)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
[∵ \(\frac{d}{d x}\) tan^-1 x = \(\frac{1}{1+x^2}\)]
= \(\frac{2}{3}\left(\tan ^{-1} x\right)^{\frac{3}{2}}\) + C

(ii) As \(\frac{d}{d x}\) sin^-1 2x = \(\frac{1}{\sqrt{1-4 x^2}}\) × 2
∴ ∫ \(\frac{\left(\sin ^{-1} 2 x\right)^3}{\sqrt{1-4 x^2}}\) dx = \(\frac{1}{2}\) ∫ (sin^-1 2x)³ \(\frac{2}{\sqrt{1-4 x^2}}\) dx
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= \(\frac{1}{2} \frac{\left(\sin ^{-1} 2 x\right)^{3+1}}{(3+1)}\) + C
= \(\frac{1}{8}\) (sin^-1 2x)⁴ + C

Question 7.
(i) ∫ sin x cos⁵ x dx
(ii) ∫ cos³ x (ax + b) sin (ax + b) dx
Solution:
(i) Let I =∫ sin x cos⁵ x dx
= – ∫ (cos x)⁵ (- sin x) dx
= – \(\frac{(\cos x)^{5+1}}{5+1}\) + C
= – \(\frac{1}{6}\) cos⁶ x + C

(ii) Let I = ∫ cos³ x (ax + b) sin (ax + b) dx
= – \(\frac{1}{a}\) ∫ (cos (ax + b))³ (- a sin (ax + b)) dx
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]
= – \(\frac{1}{a} \frac{(\cos (a x+b))^4}{4}\) + C
= – \(\frac{1}{4 a}\) cos⁴ (ax + b) + C

Question 8.
(i) ∫ sec³ x tan x dx
(ii) ∫ cot³ x cosec² x dx
Solution:
(i) Let I = ∫ sec³ x tan x dx
= ∫ sec² x (sec x tan x) dx
= \(\frac{(\sec x)^{2+1}}{2+1}\) + C
= \(\frac{\sec ^3 x}{3}\) + C

(ii) Let I = ∫ cot³ x cosec² x dx
= ∫ (cot x)³ cosec² x dx
put cos x = t
⇒ d (cot x) = dt
⇒ – cosec² x dx = dt
⇒ cosec² x dx = – dt
∴ I = ∫ t³ (- dt)
= \(\frac{-t^4}{4}\) + c
= \(\frac{- 1}{4}\) cot⁴ x + c

Question 9.
(i) ∫ \(\sqrt{tan x}\) (1 + tan² x) dx
(ii) ∫ \(\frac{\sin x}{1+\cos x}\) dx
Solution:
(i) Let I = ∫ \(\sqrt{tan x}\) (1 + tan² x) dx
= ∫ (tan x)^{\(\frac{1}{2}\)} sec² x dx
= \(\frac{(\tan x)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
= \(\frac{2}{3} \tan \frac{3}{2} x\) + C

(ii) Let I = ∫ \(\frac{\sin x}{1+\cos x}\) dx
= – ∫ \(\frac{-\sin x d x}{1+\cos x}\)
= – log (1 + cos x) + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x) + C]

Question 10.
(i) ∫ \(\frac{\cot x}{\sqrt{\sin x}}\) dx
(ii) ∫ \(\frac{\tan x}{\sqrt{\cos x}}\) dx
Solution:
(i) Let I = ∫ \(\frac{\cot x}{\sqrt{\sin x}}\) dx
= ∫ \(\frac{\cos x}{(\sin x)^{3 / 2}}\) dx
put sin x = t
⇒ cos x dx = dt
∴ I = ∫ \(\frac{d t}{t^{3 / 2}}\)
= ∫ t^-3/2 dt
= \(\frac{t^{-3 / 2+1}}{\left(\frac{-3}{2}+1\right)}\) + c
= – 2t^-1/2 + c
= \(\frac{-2}{\sqrt{\sin x}}\) + c

(ii) Let I = ∫ \(\frac{\tan x}{\sqrt{\cos x}}\) dx
= ∫ \(\frac{\sin x}{(\cos x)^{3 / 2}}\) dx
put cos x = t
⇒ d (cos x) = dt
⇒ – sin x dx = dt
⇒ sin x dx = – dt
∴ I = ∫ \(\frac{-d t}{t^{3 / 2}}\)
= – \(\frac{t^{\frac{-3}{2}+1}}{\left(\frac{-3}{2}+1\right)}\) + c
= \(\frac{2}{t^{1 / 2}}\) + c
= \(\frac{2}{\sqrt{\cos x}}\) + c

Question 11.
(i) ∫ \(\frac{\sin x}{\sqrt{3+2 \cos x}}\) dx
(ii) ∫ \(\frac{1+\sin x}{\sqrt{x-\cos x}}\) dx.
Solution:
(i) Let I = ∫ \(\frac{\sin x}{\sqrt{3+2 \cos x}}\) dx
= – \(\frac{1}{2}\) ∫ (3 + 2 cos x)^{– \(\frac{1}{2}\)} (- 2 sin x) dx
= – \(\frac{1}{2} \frac{(3+2 \cos x)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= – \(\sqrt{3+2 cos x}\) + C

(ii) Let I = ∫ \(\frac{1+\sin x}{\sqrt{x-\cos x}}\) dx
put x – cos x = t
⇒ (1 + sin x) dx = dt
∴ I = ∫ \(\frac{d t}{\sqrt{t}}\)
= ∫ t^-1/2 dt
= \(\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\) + c
= 2√t + c
= 2 \(\sqrt{x-cos x}\) + c

Question 12.
(i) ∫ \(\frac{\sin x}{(1+\cos x)^2}\) dx
(ii) ∫ \(\frac{\left(3 \tan ^2 x+2\right) \sec ^2 x}{\left(\tan ^3 x+2 \tan x+5\right)^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{\sin x}{(1+\cos x)^2}\) dx
= ∫ (1 + cos x)^-2 sin x dx
= – ∫ (1 + cos x)^-2 (- sin x dx)
= – \(\frac{(1+\cos x)^{-2+1}}{-2+1}\) + C
= \(\frac{1}{1+\cos x}\) + C

(ii) Let I = ∫ \(\frac{\left(3 \tan ^2 x+2\right) \sec ^2 x}{\left(\tan ^3 x+2 \tan x+5\right)^2}\) dx
Since \(\frac{d}{d x}\) (tan³ x + 2 tan x + 5)
= 3 tan² x sec² x + 2 sec² x
= sec² x (3 tan² x + 2)
∴ I = ∫ (tan³ x + 2 tan x + 5)^{– 2} {(3 tan² x + 2) sec² x dx}
= \(\frac{\left(\tan ^3 x+2 \tan x+5\right)^{-2+1}}{-2+1}\) + C
= – \(\frac{1}{\tan ^3 x+2 \tan x+5}\) + C
[∵ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + C, where n ≠ – 1]

Question 13.
(i) ∫ (e^3x + 1)² e^3x dx
(ii) ∫ \(\frac{e^{4 x}}{\left(1+e^{4 x}\right)^2}\) dx
Solution:
(i) Let I = ∫ (e^3x + 1)² e^3x dx
[∵ \(\frac{d}{d x}\) (e^3x + 1) = 3e^3x]
= \(\frac{1}{3}\) ∫ (e^3x + 1)² (3e^3x dx)
= \(\frac{1}{3} \frac{\left(e^{3 x}+1\right)^3}{3}\) + C
= \(\frac{1}{9}\) (e^3x + 1)³ + C

(ii) Let I = ∫ \(\frac{e^{4 x}}{\left(1+e^{4 x}\right)^2}\) dx
= ∫ (1 + e^4x)^{– 3} e^4x dx
= \(\frac{1}{4}\) ∫ (1 + e^4x)^{– 2} 4 e^4x dx
= \(\frac{1}{4} \frac{\left(1+e^{4 x}\right)^{-2+1}}{-2+1}\) + C
= – \(\frac{1}{4} \frac{1}{\left(1+e^{4 x}\right)}\) + C

Question 14.
(i) ∫ tan⁴ x dx
(ii) ∫ cot⁴ x dx
Solution:
(i) Let I = ∫ tan⁴ x dx
= ∫ tan² x . tan² x dx
= ∫ tan² x (sec² x – 1)
= ∫ tan² x sec² x dx – ∫ tan² x dx
= ∫ (tan x)² sec² x dx – ∫ (sec² x – 1) dx
= \(\frac{(\tan x)^{2+1}}{(2+1)}\) – tan x + x + C
= \(\frac{\tan ^3 x}{3}\) – tan x + x + C

(ii) Let I = ∫ cot⁴ x dx
= ∫ cot² x . cot² x dx
= ∫ cot² x (cosec² x – 1) dx
= ∫ (cot x)² cosec² x dx – ∫ (cosec² x – 1) dx
= – ∫ (cot x)² (- cosec² x dx) – ∫ cosec² x dx + ∫ dx + C
= – \(\frac{\cot ^3 x}{3}\) – cot x + x + C

Question 15.
(i) ∫ \(\frac{1+\tan x}{x+\log \sec x}\) dx
(ii) ∫ \(\frac{1+\sin 2 x}{x+\sin ^2 x}\) dx
Solution:
(i) Since \(\frac{d}{d x}\) (x + log sec x) = 1 + \(\frac{1}{\sec x}\) × sec x tan x
= 1 + tan x
∴ ∫ \(\frac{1+\tan x}{x+\log \sec x}\) dx = log |x + log sec x| + C
[∵∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

(ii) Since \(\frac{d}{d x}\) (x + sin² x) = 1 + 2 sin x cos x
= 1 + sin 2x
∴ I = ∫ \(\frac{1+\sin 2 x}{x+\sin ^2 x}\) dx
= log |x + sin² x| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

Question 16.
(i) ∫ \(\frac{\tan x \sec ^2 x}{1-\tan ^2 x}\) dx
(ii) ∫ \(\frac{x^5}{1+x^6}\) dx
Solution:
(i) Since \(\frac{d}{d x}\) (1 – tan² x) = – 2 tan x sec² x
Let I = ∫ \(\frac{\tan x \sec ^2 x}{1-\tan ^2 x}\) dx
= – \(\frac{1}{2} \int \frac{-2 \tan x \sec ^2 x d x}{1-\tan ^2 x}\)
= – \(\frac{1}{2}\) log |1 – tan² x| + C

(ii) Since \(\frac{d}{d x}\) (1 + x⁶) = 6x⁵
I = ∫ \(\frac{x^5}{1+x^6}\) dx
= ∫ \(\frac{6 x^5 d x}{1+x^6}\)
= \(\frac{1}{6}\) log |1 + x⁶| + C

Question 17.
(i) ∫ \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx
(ii) ∫ \(\frac{\sin 2 x-\sin x}{\sin ^2 x+\cos x+7}\) dx
Solution:
(i) Let I = ∫ \(\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}\) dx
= \(\frac{1}{2} \int \frac{2 \cos x-3 \sin x}{3 \cos x+2 \sin x}\) dx
put 3 cos x + 2 sin x = t
⇒ d (3 cos x + 2 sin x) = dt
⇒ (- 3 sin x + 2 cos x) dx = dt
⇒ (2 cos x – 3 sin x) dx = dt
∴ I = ∫ \(\frac{d t}{2 t}\)
= \(\frac{1}{2}\) log |t| + c
= \(\frac{1}{2}\) log |3 cos x + 2 sin x| + c

(ii) Since \(\frac{d}{d x}\) (sin² x + cos x + 7) = 2 sin x cos x – sin x
= sin 2x – sin x
∴ I = ∫ \(\frac{\sin 2 x-\sin x}{\sin ^2 x+\cos x+7}\) dx
= log |sin² x + cos x + 7| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

Question 18.
(i) ∫ \(\frac{x+\tan x}{x^2-2 \log \cos x}\) dx
(ii) ∫ \(\frac{1}{x \log x}\) dx
Solution:
(i) Since \(\frac{d}{d x}\) (x² – 2 lo0g cos x) = 2x – \(\frac{2}{\cos x}\) (- sin x)
= 2x + 2 tan x
Let I = ∫ \(\frac{(x+\tan x) d x}{x^2-2 \log \cos x}\)
= \(\frac{1}{2}\) ∫ \(\frac{2(x+\tan x) d x}{x^2-2 \log \cos x}\)
= \(\frac{1}{2}\) log |x² – 2 log cos x| + C

(ii) Let I = ∫ \(\frac{d x}{x \log x}\) ;
put log x = t
⇒ \(\frac{1}{x}\) dx = dt
∴ I = ∫ \(\frac{d t}{t}\)
= log |t| + c
= log |log x| + c

Question 19.
(i) ∫ \(\frac{2 x^2-1}{x\left(x^2-\log x\right)}\) dx
(ii) ∫ \(\frac{e^{2 x}+1}{e^{2 x}-1}\) dx
Solution:
(i) Let I = ∫ \(\frac{2 x^2-1}{x\left(x^2-\log x\right)}\) dx
= \(\frac{\frac{2 x^2-1}{x}}{x^2-\log x}\) dx
= \(\frac{\left(2 x-\frac{1}{x}\right) d x}{x^2-\log x}\) dx
As \(\frac{d}{d x}\) (x² – log x) = 2x – \(\frac{1}{x}\)
I = log |x² – log x| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

(ii) Let I = ∫ \(\frac{e^{2 x}+1}{e^{2 x}-1}\) dx
Divivde Num and Deno. by e^x ; we have
= ∫ \(\frac{e^x+e^{-x}}{e^x-e^{-x}}\) dx
put e^x – e^{– x} = t
⇒ (e^x + e^{– x}) dx = dt
∴ I = ∫ \(\frac{d t}{t}\)
= log |e^x – e^{– x}| + C.

Question 20.
(i) ∫ \(\frac{1}{1+e^{-x}}\) dx
(ii) ∫ \(\frac{\tan x}{a+b \tan ^2 x}\) dx
Solution:
(i) Let I = ∫ \(\frac{1}{1+e^{-x}}\) dx
= ∫ \(\frac{e^x d x}{1+e^x}\)
= log |1 + e^x| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

(ii) Let I = ∫ \(\frac{\tan x}{a+b \tan ^2 x}\) dx
= ∫ \(\frac{\frac{\sin x d x}{\cos x}}{a+b \frac{\sin ^2 x}{\cos ^2 x}}\)
= ∫ \(\frac{\sin x \cos x d x}{a \cos ^2 x+b \sin ^2 x}\)
Since, \(\frac{d}{d x}\) (a cos² x + b sin² x)
= 2a cos x (- sin x) + 2b sin x cos x
= 2 (b – a) cos x sin x
∴ I = \(\frac{1}{2(b-a)} \int \frac{2(b-a) \sin x \cos x d x}{a \cos ^2 x+b \sin ^2 x}\)
= \(\frac{1}{2(b-a)}\) log |a cos² x + b sin² x| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

Question 21.
(i) ∫ \(\frac{1-\cot x}{1+\cot x}\) dx
(ii) ∫ \(\frac{\cos 2 x}{(\cos x+\sin x)^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{1-\cot x}{1+\cot x}\) dx
= ∫ \(\frac{1-\frac{\cos x}{\sin x}}{1-\frac{\cos x}{\sin x}}\) dx
= ∫ \(\frac{(\sin x-\cos x)}{\sin x+\cos x}\) dx
putting sin x + cos x = t
⇒ d (sin x + cos x) = dt
⇒ (cos x – sin x) dx = dt
⇒ (sin x – cos x) dx = – dt
∴ I = ∫ \(-\frac{d t}{t}\)
= – log |t| + c
= – log |cos x + sin x| + c

(ii) Let I = ∫ \(\frac{\cos 2 x}{(\cos x+\sin x)^2}\) dx
Thus I = ∫ \(\frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2}\) dx
= ∫ \(\frac{\cos x-\sin x}{\cos x+\sin x}\) dx
put cos x + sin x = t
⇒ d (cos x + sin x) = dt
⇒ (- sin x + cos x) dx = dt
∴ I = ∫ \(\frac{d t}{t}\)
= log |t| + c
= log |sin x + cos x| + c

After:
I = ∫ \(\frac{\cos 2 x d x}{\cos ^2 x+\sin ^2 x+2 \sin x \cos x}\)
= \(\frac{\cos 2 x d x}{1+\sin 2 x}\)
= \(\frac{2 \cos 2 x d x}{1+\sin 2 x}\)
= \(\frac{1}{2}\) log |1 + sin 2x| + c
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) = log |f(x)| + C]

Peer review of Class 12 ISC Maths Solutions Chapter 8 Integrals Ex 8.3 can encourage collaborative learning.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.3

Very short answer type questions (1 to 10):

Find all the anti-derivatives of the following (1 to 6) functions :

Question 1.
(i) (ax + b)³
(ii) 4e^3x + 1 (NCERT)
Solution:
(i) ∫ (ax + b)³ = \(\frac{(a x+b)^{3+1}}{(3+1) a}\) + C
= \(\frac{(a x+b)^4}{4 a}\) + C
[∵ ∫ (ax + b)ⁿ dx = \(\frac{(a x+b)^{n+1}}{(n+1) a}\) + C ; n ≠ – 1]

(ii) ∫ 4e^3x + 1 dx
= 4 ∫ e^3x dx + ∫ dx
= \(\frac{4 e^{3 x}}{3}\) + x + C

Question 1 (old).
(i) sin 2x (NCERT)
(ii) cos 3x (NCERT)
Solution:
(i) ∫ sin 2x dx = – \(\frac{\cos 2 x}{2}\) + C
[∵ ∫ sin mx dx = \(\frac{\cos m x}{m}\) + C]

(ii) ∫ cos 3x dx = \(\frac{\sin 3 x}{3}\) + C

Question 2.
(i) \(\frac{1}{\sqrt{7-2 x}}\)
(ii) \(\sqrt{a x+b}\) (NCERT)
Solution:
(i) ∫ \(\frac{1}{\sqrt{7-2 x}}\) dx
= ∫ (7 – 2x)^{– \(\frac{1}{2}\)} dx
= \(\frac{(7-2 x)^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)(-2)}\) + C
= – \(\sqrt{7-2x}\) + C

(ii) ∫ \(\sqrt{a x+b}\) dx
= ∫ (ax + b)^1/2 dx
= \(\frac{(a x+b)^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right) a}\)
= \(\frac{2}{3 a}\) (ax + b)^3/2 + C

Question 2 (old).
(i) sin mx (NCERT)
(ii) e^2x (NCERT)
Solution:
(i) ∫ sin mx dx = – \(\frac{\cos m x}{m}\) + C

(ii) ∫ e^2x dx
= \(\frac{e^{2 x}}{2}\) + C
[∵ ∫ e^mx dx = \(\frac{e^{m x}}{m}\) + C]

Question 3.
(i) tan² (2x – 3) (NCERT)
(ii) 5 cosec² (2 – 7x)
Solution:
(i) ∫ tan² (2x – 3) dx
= ∫ [sec² (2x – 3) – 1] dx
= ∫ [sec² (2x – 3) dx – ∫ dx
= \(\frac{\tan (2 x-3)}{2}\) – x + c

(ii) ∫ 5 cosec² (2 – 7x) dx
= \(\frac{-5 \cot (2-7 x)}{-7}\) + C
= \(\frac{5}{7}\) cot (2 – 7x) + C

Question 4.
(i) \(\sqrt{e^x}\)
(ii) cosec (3x + 2) cot (3x + 2)
Solution:
(i) ∫ \(\sqrt{e^x}\) dx
= ∫ e^x/2 dx
= \(\frac{e^{x / 2}}{1 / 2}\) + C
= 2e^x/2 + C

(ii) ∫ cos (3x + 2) cot (3x + 2) dx
= – \(\frac{\ {cosec}(3 x+2)}{3}\) + C
[∵ ∫ cot θ cosec θ dθ = – cosec θ + C]

Evaluate the following (7 to 21) integrals :

Question 5.
(i) ∫ e^{2x + 3} dx
(ii) ∫ e^{(3a log x + e3x log a)} dx
Solution:
(i) ∫ e^{2x + 3} dx
= \(\frac{e^{2 x+3}}{2}\) + C
[∵ ∫ e^mx dx = \(\frac{e^{m x}}{m}\) + C]

(ii) ∫ e^{(3a log x + e3x log a)} dx
= ∫ (e^{log x3a} + e^{log a3x}) dx
[∵ a log b = log b^a
and e^{log x} = x]
= ∫ (x^3a + a) dx
= \(\frac{x^{3 a+1}}{3 a+1}+\frac{a^{3 x}}{3 \log a}\) + C

Question 6.
(i) ∫ \(\sqrt{1+cos x}\) dx
(ii) ∫ \(\sqrt{1+sin x}\) dx
Solution:
(i) ∫ \(\sqrt{1+cos x}\) dx
= ∫ \(\sqrt{2 \cos ^2 \frac{x}{2}}\) dx
= √2 ∫ cos \(\frac{x}{2}\) dx
= √2 \(\frac{\sin \frac{x}{2}}{\frac{1}{2}}\) + C
= 2√2 sin \(\frac{x}{2}\) + C

(ii) ∫ \(\sqrt{1+sin x}\) dx

Question 6 (old).
(ii) ∫ \(\sqrt{1-sin x}\) dx
Solution:

Question 7.
(i) ∫ sec² (7 – x) dx
(ii) ∫ a^{3x + 5} dx, a ≥ 0
Solution:
(i) ∫ sec² (7 – x) dx
= ∫ sec² t (- dt)
[put 7 – x = t
⇒ dx = – dt]
= – tan t + C
= – tan (7 – x) + C

(ii) ∫ a^{3x + 5} dx, a > 0
= \(\frac{a^{3 x+5}}{3(\log a)}\) + C
[∵ ∫ a^mx dx = \(\frac{a^{m x}}{m \log a}\) + C

Question 8.
(i) ∫ \(\frac{1-\tan ^2 x}{1+\tan ^2 x}\) dx
(ii) ∫ sec² x cosec² x dx
Solution:
(i) ∫ \(\frac{1-\tan ^2 x}{1+\tan ^2 x}\) dx
= ∫ cos 2x dx
= \(\frac{\sin 2 x}{2}\) + C

(ii) ∫ sec² x cosec² x dx
= ∫ \(\frac{\left(\sin ^2 x+\cos ^2 x\right) d x}{\sin ^2 x \cos ^2 x}\)
= ∫ \(\frac{1}{\cos ^2 x}\) dx + ∫ \(\frac{1}{\sin ^2 x}\) dx
= ∫ sec² x dx + ∫ cosec² x dx
= tan x – cot x + C

Question 9.
(i) ∫ \(\frac{x}{2 x-3}\) dx
(ii) ∫ \(\frac{x+2}{3 x^2+5 x-2}\) dx
Solution:
(i) Let I = ∫ \(\frac{x}{2 x-3}\) dx
= \(\frac{1}{2} \int\left[\frac{2 x-3+3}{2 x-3}\right]\) dx
= \(\frac{1}{2} \int\left[1+\frac{3}{2 x-3}\right]\) dx
= \(\frac{1}{2}\left[x-\frac{3 \log |2 x-3|}{2}\right]\) + C
= \(\frac{x}{2}\) – \(\frac{3}{4}\) log |2x – 3| + C

(ii) Let I = ∫ \(\frac{x+2}{3 x^2+5 x-2}\) dx
= ∫ \(\frac{(x+2) d x}{(x+2)(3 x-1)}\)
= ∫ \(\frac{d x}{3 x-1}\)
= \(\frac{\log |3 x-1|}{3}\) + C

Question 10.
(i) ∫ \(\frac{x}{(x+1)^2}\) dx (ISC 2010)
(ii) ∫ \(\frac{x}{(x+1)^2}\) dx (NCERT Exemplar)
Solution:
(i) ∫ \(\frac{x}{(x+1)^2}\) dx

(ii) ∫ \(\frac{x}{(x+1)^2}\) dx
= ∫ \(\left[\frac{x^2-1+3}{x+1}\right]\) dx
= ∫ [(x – 1) + \(\frac{3}{x+1}\)] dx
= \(\frac{x^{2}}{2}\) – x + 3 log |x + 1| + C

Question 11.
(i) ∫ \(\frac{x^3}{2 x+1}\) dx
(ii) ∫ \(\frac{x^3+4 x^2-3 x-2}{x+2}\) dx
Solution:
(i) ∫ \(\frac{x^3}{2 x+1}\) dx

(ii) ∫ \(\frac{x^3+4 x^2-3 x-2}{x+2}\) dx
= ∫ [(x² + 2x – 7) + \(\frac{12}{x+2}\)] dx
= \(\frac{x^3}{3}\) + x² – 7x + 12 log |x + 2| + C

Question 11 (old).
(i) ∫ \(\frac{1}{(2 x-3)^{3 / 2}}\) dx
(ii) ∫ \(\frac{x}{2 x-3}\) dx
Solution:
(i) ∫ \(\frac{1}{(2 x-3)^{3 / 2}}\) dx
= ∫ (2x – 3)^{– 3/2} dx
= \(\frac{(2 x-3)^{-\frac{3}{2}+1}}{\left(-\frac{3}{2}+1\right) \cdot 2}\)
= – \(\frac{1}{\sqrt{2 x-3}}\) + C

(ii) ∫ \(\frac{x}{2 x-3}\) dx
= \(\frac{1}{2} \int\left[\frac{2 x-3+3}{2 x-3}\right]\) dx
= \(\frac{1}{2} \int\left[1+\frac{3}{2 x-3}\right]\) dx
= \(\frac{1}{2}\left[x-\frac{3 \log |2 x-3|}{2}\right]\) + C
= \(\frac{x}{2}\) – \(\frac{3}{4}\) log |2x – 3| + C

Question 12.
(i) ∫ \(\frac{1}{\sqrt{x+1}+\sqrt{x+2}}\) dx
(ii) ∫ \(\frac{1}{\sqrt{2 x+3}-\sqrt{2 x}}\) dx
Solution:
(i) ∫ \(\frac{1}{\sqrt{x+1}+\sqrt{x+2}}\) dx
= ∫ \(\frac{\sqrt{x+1}-\sqrt{x+2}}{(x+1)-(x+2)}\) dx
= – \(\left[\frac{2}{3}(x+1)^{3 / 2}-\frac{2}{3}(x+2)^{3 / 2}\right]\) + C

(ii) ∫ \(\frac{1}{\sqrt{2 x+3}-\sqrt{2 x}}\) dx

Question 13.
(i) ∫ \(\frac{x+1}{\sqrt{2 x-1}}\) dx
(ii) ∫ x \(\sqrt{3 x-2}\) dx
Solution:
(i) ∫ \(\frac{x+1}{\sqrt{2 x-1}}\) dx

(ii) ∫ x \(\sqrt{3 x-2}\) dx
= \(\frac{1}{3}\) ∫ (3x – 2 + 2) \(\sqrt{3 x-2}\) dx
= \(\frac{1}{3}\) ∫ (3x – 2)^3/2 dx + \(\frac{2}{3}\) ∫ (3x – 2)^1/2 dx
= \(\frac{1}{3} \frac{(3 x-2)^{5 / 2}}{\frac{5}{2} \times 3}+\frac{2}{3} \frac{(3 x-2)^{3 / 2}}{\frac{3}{2} \times 3}\) + C
= \(\frac{2}{45}\) (3x – 2)^5/2 + \(\frac{4}{27}\) (3x – 2)^3/2 + C

Question 14.
(i) ∫ cos² x dx (NCERT)
(ii) ∫ sin⁴ dx
Solution:
(i) ∫ cos² x dx
= ∫ \(\frac{1+\cos 2 x}{2}\) dx
= \(\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]\) + C

(ii) ∫ sin⁴ dx
= ∫ \(\left[\frac{1-\cos 2 x}{2}\right]^2\) dx
= \(\frac{1}{4}\) ∫ [1 + cos² 2x – 2 cos 2x] dx
= \(\frac{1}{4}\) ∫ [1 + \(\frac{1+\cos 4 x}{2}\) – 2 cos 2x] dx
= \(\frac{1}{8}\) ∫ [3 + cos 4x – 4 cos 2x] dx
= \(\frac{1}{8}\) [3x + \(\frac{\sin 4 x}{4}\) – 2 sin 2x] + C

Question 15.
(i) ∫ sin⁵ (2x + 5) dx (NCERT)
(ii) ∫ cos⁴ 2x dx (NCERT)
Solution:
(i) ∫ sin⁵ (2x + 5) dx
= ∫ \(\frac{1-\cos (4 x+10)}{2}\) dx
[∵ sin² θ = \(\frac{1-\cos 2 \theta}{2}\)]
= ∫ [x – \(\frac{\sin (4 x+10)}{4}\)] + C

(ii) ∫ cos⁴ 2x dx
= ∫ (cos² 2x)² dx
= ∫ \(\left[\frac{1+\cos 4 x}{2}\right]^2\) dx
= \(\frac{1}{4}\) ∫ [1 + 2 cos 4x + cos² 4x] dx
= \(\frac{1}{4}\) ∫ [1 + 2 cos 4x + \(\frac{1+\cos 8 x}{2}\)] dx
= \(\frac{1}{8}\) ∫ [3 + 4 cos 4x + cos 8x] dx
= \(\frac{1}{8}\) [3x + 4 \(\frac{\sin 4 x}{4}\) + \(\frac{\sin 8 x}{8}\)] + c
= \(\frac{3 x}{8}+\frac{\sin 4 x}{8}+\frac{\sin 8 x}{64}\) + c

Question 16.
(i) ∫ \(\frac{1}{1-\sin \frac{x}{2}}\) dx
(ii) ∫ sin x \(\sqrt{1-cos 2x}\) dx
Solution:
(i) ∫ \(\frac{1}{1-\sin \frac{x}{2}}\) dx

(ii) ∫ sin x \(\sqrt{1-cos 2x}\) dx
= ∫ sin x \(\sqrt{2 \sin ^2 x}\) dx
= √2 ∫ sin² x dx
= \(\sqrt{2} \int \frac{1-\cos 2 x}{2}\) dx
= \(\frac{1}{\sqrt{2}}\left[x-\frac{\sin 2 x}{2}\right]\) + C
= \(\frac{x}{\sqrt{2}}\) – \(\frac{1}{2 \sqrt{2}}\) sin 2x + C.

Question 17.
(i) ∫ cos 3x cos 5x dx
(ii) ∫ sin 4x sin 8x dx
(iii) ∫ sin 4x cos 3x dx
(iv) ∫ sin 2x cos 3x dx (NCERT)
Solution:
(i) ∫ cos 3x cos 5x dx
= \(\frac{1}{2}\) ∫ (2 cos 5x cos 3x) dx
= \(\frac{1}{2}\) ∫ [cos 8x + cos 2x] dx
= \(\frac{1}{2}\left[\frac{\sin 8 x}{8}+\frac{\sin 2 x}{2}\right]\) + C

(ii) ∫ sin 4x sin 8x dx
= \(\frac{1}{2}\) ∫ (2 sin 8x sin 4x) dx
= \(\frac{1}{2}\) ∫ [cos 4x – cos 12 x] dx
[∵ 2 sin A sin B = cos (A – B) – cos (A + B)]
= \(\frac{1}{2}\left[\frac{\sin 4 x}{7}-\frac{\sin 12 x}{12}\right]\) + C

(iii) ∫ sin 4x cos 3x dx
= \(\frac{1}{2}\) ∫ 2 sin 4x cos 3x dx
= \(\frac{1}{2}\) ∫ (sin 7x + sin x) dx
[∵ 2 sin A cos B = sin (A + B) + sin (A – B)]
= \(\frac{1}{2}\left[-\frac{\cos 7 x}{7}-\cos x\right]\) + C
= \(\frac{1}{14}\) [- cos 7x – 7 cos x] + C

(iv) ∫ sin 2x cos 3x dx
= \(\frac{1}{2}\) ∫ (2 sin 2x cos 3x) dx
= \(\frac{1}{2}\) ∫ [sin 5x + sin (- x)] dx
= \(\frac{1}{2}\) ∫ (sin 5x – sin x) dx
[∵ 2 sin A cos B = sin (A + B) + sin (A – B)]
= \(\frac{1}{2}\) [\(\frac{-\cos 5 x}{5}\) + cos x] + C

Question 18.
(i) ∫ \(\frac{\sin 4 x}{\cos x}\) dx
(ii) ∫ sin x sin 2x sin 3x dx
Solution:
(i) ∫ \(\frac{\sin 4 x}{\cos x}\) dx
= ∫ \(\frac{2 \sin 2 x \cos 2 x}{\cos x}\) dx
= ∫ \(\frac{2 \times 2 \sin x \cos x \cos 2 x d x}{\cos x}\) dx
= 2 ∫ 2 cos 2x sin x dx
= 2 ∫ [sin 3x – sin x] dx
= 2 [- \(\frac{\cos 3 x}{3}\) + cos x]
= \(\frac{2}{3}\) (3 cos x – cos 3x)

(ii) ∫ sin x sin 2x sin 3x dx
= \(\frac{1}{2}\) ∫ (2 sin 3x sin 2x) sin x dx
= \(\frac{1}{2}\) ∫ [cos x – cos 5x] sin x dx
= \(\frac{1}{4}\) ∫ 2 sin x cos x dx – \(\frac{1}{4}\) ∫ 2 cos 5x sin x dx
= \(\frac{1}{4}\left(-\frac{\cos 2 x}{2}\right)\) – \(\frac{1}{4}\) ∫ [sin 6x – sin 4x] dx
= \(-\frac{\cos 2 x}{8}+\frac{\cos 6 x}{24}-\frac{\cos 4 x}{16}\) + C

Question 19.
(i) ∫ sin⁶ x dx
(ii) ∫ sin⁴ x cos² x dx.
Solution:
(i) ∫ sin⁶ x dx
= ∫ (sin² x)³ dx
= ∫ \(\left[\frac{1-\cos 2 x}{2}\right]^3\) dx
= \(\frac{1}{8}\) ∫ [1 – \(\frac{1}{4}\) (cos 6x + 3 cos 2x) – 3 cos 2x + 3 \(\left(\frac{1+\cos 4 x}{2}\right)\)]
[∵ cos³ θ = \(\frac{1}{4}\) [cos 3θ + 3 cos θ]
and cos² θ = \(\frac{1+\cos 2 \theta}{2}\)]
= \(\frac{1}{8 \times 4}\) ∫ [4 – cos 6x – 3 cos 2x – 12 cos 2x + 6 + 6 cos 4x] dx
= \(\frac{1}{32}\) ∫ [10 + 6 cos 4x – 15 cos 2x – cos 6x]
= \(\frac{1}{32}\left[10 x+\frac{6 \sin 4 x}{4}-\frac{15 \sin 2 x}{2}-\frac{\sin 6 x}{6}\right]\) + C
= \(\frac{1}{192}\) [60x + 9 sin 4x – 45 sin 2x – sin 6x] + C

(ii) ∫ sin⁴ x cos² x dx.
= \(\frac{1}{4}\) ∫ sin² x (4 sin² x cos² x) dx
= \(\frac{1}{4}\) ∫ \(\left[\frac{1-\cos 2 x}{2}\right]\) [2 sin x cos x]² dx
= \(\frac{1}{4} \int\left[\frac{1-\cos 2 x}{2}\right]\left[\frac{1-\cos 4 x}{2}\right]\) dx
= \(\frac{1}{16}\) ∫ [1 – cos 4x – cos 2x + (2 cos 4x cos 2x) \(\frac{1}{2}\)] dx
= \(\frac{1}{16}\) ∫ [1 – cos 4x – cos 2x – \(\frac{1}{2}\) (cos 6x + cos 2x)] dx
= \(\frac{1}{32}\) ∫ [2 – 2 cos 4x – cos 2x + cos 6x] dx
= \(\frac{1}{32}\left[2 x-\frac{2 \sin 4 x}{4}-\frac{\sin 2 x}{2}+\frac{\sin 6 x}{6}\right]\) + C
= \(\frac{1}{192}\) [12x – 3 sin 4x – 3 sin 2x + sin 6x] + C

Question 20.
Evaluate ∫ cos mx cos nx dx, where m, n are positive integers, m ≠ n. What happens if m n?
Solution:
Case – I:
When m ≠ n
∴ ∫ cos mx cos nx dx = ∫ 2 cos mx cos nx dx
= \(\frac{1}{2}\) ∫ [cos (m + n) x + cos (m – n) dx
= \(\frac{1}{2}\left[\frac{\sin (m+n) x}{m+n}+\frac{\sin (m-n) x}{m-n}\right]\) + C

Case-II:
When m = n
∫ cos mx cos nx dx = ∫ cos² mx dx
= ∫ \(\left[\frac{1+\cos 2 m x}{2}\right]\) dx
= \(\frac{1}{2}\left[x+\frac{\sin 2 m x}{2 m}\right]\) + C

Continuous practice using ISC Maths Class 12 Solutions Chapter 8 Integrals Ex 8.2 can lead to a stronger grasp of mathematical concepts.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.2

Very short answer type questions (1 to 14) :

Evaluate the following (1 to 20) integrals :

Question 1.
(i) ∫ (ax² + bx + c) dx (NCERT)
(ii) ∫ (x^2/3 + 1) dx. (NCFRT)
Solution:
(i) ∫ (ax² + bx + c) dx
= a ∫ x² dx + b ∫ x dx + ∫ c dx
= \(\frac{a x^3}{3}+\frac{b x^2}{2}\) + cx + C
[∵ ∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C; n ≠ 1]

(ii) ∫ (x^2/3 + 1) dx
= ∫ x^2/3 dx + ∫ 1 dx
= \(\frac{x^{\frac{2}{3}+1}}{\frac{2}{3}+1}\) + x + C
= \(\frac{3}{5}\) x^5/3 + x + C

Question 2.
(i) ∫ (\(\sqrt{3x}\) + \(\frac{1}{\sqrt{x}}\)) dx
(ii) ∫ \(\left(\frac{2 a}{\sqrt{x}}-\frac{b}{x^2}+3 c \sqrt[3]{x^2}\right)\) dx (NCERT Exampler)
Solution:
(i) ∫ (\(\sqrt{3x}\) + \(\frac{1}{\sqrt{x}}\)) dx
= ∫ \(\sqrt{3x}\) dx + ∫ \(\frac{1}{\sqrt{x}}\)) dx
= √3 ∫ x^{\(\frac{1}{2}\)} dx + ∫ x^{\(-\frac{1}{2}\)} dx
= √3 \(\frac{x^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= \(\frac{2}{\sqrt{3}} x^{3 / 2}\) + 2√x + C

(ii) ∫ \(\left(\frac{2 a}{\sqrt{x}}-\frac{b}{x^2}+3 c \sqrt[3]{x^2}\right)\) dx
= ∫ \(\frac{2 a}{\sqrt{x}}\) dx – b ∫ x^{– 2} dx + 3c ∫ (x²)^1/3 dx
= 2a ∫ x^{\(-\frac{1}{2}\)} dx – b ∫ x^{– 2} dx + 3c ∫ x^2/3 dx
= 2a \(\frac{x^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)}-b \frac{x^{-2+1}}{(-2+1)}+3 c \frac{x^{\frac{2}{3}+1}}{\left(\frac{2}{3}+1\right)}\) + C
= 4a√x + \(\frac{b}{x}\) + \(\frac{9 c}{5}\) x^5/3 + C.

Question 3.
(i) ∫ \(\frac{2 \cos x}{3 \sin ^2 x}\) dx
(ii) ∫ (2x² 3 sin x + 5√x) dx (NCERT)
Solution:
(i) ∫ \(\frac{2 \cos x}{3 \sin ^2 x}\) dx
= \(\frac{2}{3}\) ∫ cot x cosec x dx
= – \(\frac{2}{3}\) cosec x + C

(ii) ∫ (2x² – 3 sin x + 5√x) dx
= 2 ∫ x² dx – 3 ∫ sin x dx + 5 ∫ √x dx
= 2 \(\frac{x^{3}}{3}\) – 3 (- cos x) + 5 \(\frac{x^{3 / 2}}{3 / 2}\) dx
= \(\frac{2}{3}\) x³ + 3 cos x + \(\frac{10}{3}\) x^3/2 + C

Question 4.
(i) ∫ (2x – 3 cos x + e^x) dx (NCERT)
(ii) ∫ x² (1 – \(\frac{1}{x^2}\)) dx (NCERT)
Solution:
(i) ∫ (2x – 3 cos x + e^x) dx
= ∫ 2x dx + 3 ∫ cos x dx + ∫ e^x dx
= \(\frac{2 x^2}{2}\) – 3 sin x + e^x + C
= x² – 3 sin x + e^x + C

(ii) ∫ x² (1 – \(\frac{1}{x^2}\)) dx
= ∫ x² dx – ∫ 1 dx
= \(\frac{x^{3}}{3}\) – x + C

Question 5.
(i) ∫ (x^3/2 + 2 e^x – \(\frac{1}{x}\)) dx (NCERT)
(ii) ∫ (√x + \(\frac{1}{\sqrt{x}}\)) dx (NCERT)
Solution:
(i) ∫ (x^3/2 + 2 e^x – \(\frac{1}{x}\)) dx
= \(\frac{x^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}\) + 2 e^x – log |x| + C
= \(\frac{2}{5}\) x^5/2 + 2 e^x – log |x| + C
[∵ ∫ \(\frac{d x}{x}\) = log |x| + C ;
∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) ; n ≠ – 1]

(ii) ∫ (√x + \(\frac{1}{\sqrt{x}}\)) dx
= ∫ √x dx + ∫ \(\frac{d x}{\sqrt{x}}\)
= \(\frac{x^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= \(\frac{2}{3}\) x^3/2 + 2 √x + C

Question 6.
(i) ∫ (1 – x) √x dx
(ii) ∫ √x (3x² + 2x + 3) dx (NCERT)
Solution:
(i) ∫ (1 – x) √x dx
= ∫ √x dx + ∫ x^3/2 dx
= \(\frac{x^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}-\frac{x^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}\) + C
= \(\frac{2}{3} x^{3 / 2}-\frac{2}{5} x^{5 / 2}\) + C

(ii) ∫ √x (3x² + 2x + 3) dx
= ∫ 3x^5/2 dx + ∫ 2x^3/2 dx + ∫ 3x^1/2 dx
= \(\frac{3 x^{\frac{5}{2}+1}}{\frac{5}{2}+1}+\frac{2 x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+\frac{3 x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) + C
= \(\frac{6}{7} x^{7 / 2}+\frac{4}{5} x^{5 / 2}\) + 2x^3/2 + C

Question 7.
(i) ∫ \(\frac{x^3+5 x^2+4 x+1}{x^2}\) dx (ISC 2018)
(ii) ∫ \(\frac{x^3+3 x+4}{\sqrt{x}}\) dx (NCERT)
Solution:
(i) ∫ \(\frac{x^3+5 x^2+4 x+1}{x^2}\) dx
= ∫ \(\left[\frac{x^3}{x^2}+\frac{5 x^2}{x^2}+\frac{4 x}{x^2}+\frac{1}{x^2}\right]\) dx
= ∫ x dx + ∫ 5 dx + 4 ∫ \(\frac{d x}{x}\) + ∫ x^-2 dx
= \(\frac{x^{2}}{2}\) + 5x + 4 log |x| + \(\frac{x^{-2+1}}{-2+1}\) + C
= \(\frac{x^{2}}{2}\) + 5x + 4 log |x| – \(\frac{1}{x}\) + C

(ii) ∫ \(\frac{x^3+3 x+4}{\sqrt{x}}\) dx
= ∫ x^5/2 dx + 3 ∫ √x dx + 4 ∫ x^{\(-\frac{1}{2}\)} dx
= \(\frac{x^{7 / 2}}{\frac{7}{2}}+\frac{3 x^{3 / 2}}{\frac{3}{2}}+\frac{4 x^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)}\) + C
= \(\frac{2}{7} x^{\frac{7}{2}}\) + 2x^3/2 + 8√x + C

Question 8.
(i) ∫ x² (2x – \(\frac{1}{x}\))² dx
(ii) ∫ (x + √x)³ dx
Solution:
(i) ∫ x² (2x – \(\frac{1}{x}\))² dx
= ∫ x² [4x² + \(\frac{1}{x^{2}}\) – 4] dx
= ∫ 4x⁴ dx + ∫ dx – 4 ∫ x² dx
= \(\frac{4 x^5}{5}+x-\frac{4 x^3}{3}\) + C

(ii) ∫ (x + √x)³ dx
= ∫ [x³ + x^3/2 + 3x^3/2 (x + √x)] dx
= ∫ x³ dx + ∫ x^3/2 dx + 3 ∫ x^5/2 dx + 3 ∫ x² dx
= \(\frac{x^4}{4}+\frac{2 x^{5 / 2}}{5}+\frac{3 x^{7 / 2}}{\frac{7}{2}}+\frac{3 x^3}{3}\) + C
= \(\frac{x^4}{4}+\frac{2}{5} x^{5 / 2}+\frac{6}{7} x^{7 / 2}\) + x³ + C

Question 9.
(i) ∫ \(\frac{(x+a)^2}{\sqrt{x}}\) dx
(ii) ∫ \(\frac{x^3-x^2+x-1}{x-1}\) dx
Solution:
(i) ∫ \(\frac{(x+a)^2}{\sqrt{x}}\) dx
= ∫ \(\frac{x^2+a^2+2 a x}{\sqrt{x}}\) dx
= ∫ x^3/2 dx + a² ∫ x^{– \(\frac{1}{2}\)} dx + 2a ∫ √x dx
= \(\frac{2}{5}\) x^5/2 + 2a²√x + \(\frac{4 a}{3}\) x^3/2 + C

(ii) ∫ \(\frac{x^3-x^2+x-1}{x-1}\) dx
= ∫ \(\frac{x^3-x^2+x-1}{x-1}\) dx
= ∫ \(\frac{(x-1)\left(x^2+1\right)}{(x-1)}\) dx
= ∫ (x² + 1) dx
= \(\frac{x^{3}}{3}\) + x + C

Question 10.
(i) ∫ \(\frac{1-\sin x}{\cos ^2 x}\) dx (NCERT)
(ii) ∫ \(\frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\) dx (NCERT Exemplar)
Solution:
(i) ∫ \(\frac{1-\sin x}{\cos ^2 x}\) dx
= \(\int \frac{1}{\cos ^2 x} d x-\int \frac{\sin x}{\cos ^2 x} d x\)
= ∫ sec² x dx – ∫ tan x sec x dx
= tan x – sec x + C

(ii) Let I = ∫ \(\frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\) dx
= ∫ \(\left(\frac{e^{\log x^6}-e^{\log x^5}}{e^{\log x^4}-e^{\log x^3}}\right)\) dx
= ∫ \(\frac{x^6-x^5}{x^4-x^3}\) dx
= = ∫ \(\frac{x^5(x-1)}{x^3(x-1)}\) dx
= \(\frac{x^{3}}{3}\) + C

Question 11.
(i) ∫ sec x (sec x + tan x) dx
(ii) ∫ \(\frac{\sec ^2 x}{\ {cosec}^2 x}\) dx
Solution:
(i) ∫ sec x (sec x + tan x) dx
= ∫ sec² x dx + ∫ sec x tan x dx
= tan x + sec x + C

(ii) ∫ \(\frac{\sec ^2 x}{\ {cosec}^2 x}\) dx
= ∫ \(\) × sin² x dx
= ∫ tan² x dx
= ∫ (sec² x – 1) dx
= tan x – x + C

Question 12.
(i) ∫ \(\left(\frac{\cot x}{\sin x}-\frac{\tan x}{\cos x}-3 \tan ^2 x\right)\) dx
(ii) ∫ \(\frac{2-3 \sin x}{\cos ^2 x}\) dx
Solution:
(i) ∫ \(\left(\frac{\cot x}{\sin x}-\frac{\tan x}{\cos x}-3 \tan ^2 x\right)\) dx
= ∫ cot x cosec x dx – ∫ tan x sec x dx – 3 ∫ (sec² x – 1) dx
= – cosec x – sec x – 3 tan x + 3x + C
[∵ ∫ cot x cosec x dx = – cosec x + C
and ∫ tan x sec x dx = sec x + C]

(ii) ∫ \(\frac{2-3 \sin x}{\cos ^2 x}\) dx
= ∫ 2 sec² x dx – 3 ∫ tan x sec x dx
= 2 tan x – 3 sec x + C

Question 13.
(i) ∫ \(\frac{1}{1+\cos x}\) dx
(ii) ∫ \(\frac{1}{1+\sin x}\) dx
Solution:
(i) ∫ \(\frac{1}{1+\cos x}\) dx
= ∫ \(\frac{1}{1+\cos x} \times \frac{1-\cos x}{1-\cos x}\)
= ∫ \(\frac{1-\cos x}{\sin ^2 x}\) dx
= ∫ cosec² dx – ∫ cot x cosec x dx
= – cot x + cosec x + C

(ii) ∫ \(\frac{1}{1+\sin x}\) dx
= ∫ \(\frac{(1+\sin x) d x}{1-\sin ^2 x}\)
= ∫ \(\frac{(1+\sin x)}{\cos ^2 x}\) dx
= ∫ sec^{2</sup x dx + ∫ tan x sec x dx
= tan x + sec x + C}

Question 14.
(i) ∫ \(\frac{1}{1+\ {cosec} x}\) dx
(ii) ∫ \(\frac{\sin ^2 x}{1+\cos x}\) dx (NCERT)
Solution:
(i) ∫ \(\frac{1}{1+\ {cosec} x}\) dx

(ii) ∫ \(\frac{\sin ^2 x}{1+\cos x}\) dx
= ∫ \(\frac{1-\cos ^2 x}{1+\cos x}\) dx
= ∫ \(\frac{(1-\cos x)(1+\cos x)}{1+\cos x}\) dx
= ∫ (1 – cos x) dx
= ∫ dx – ∫ cos x dx
= x – sin x + c

Question 15.
(i) ∫ (√x – sin \(\frac{x}{2}\) cos \(\frac{x}{2}\) + 5) dx
(ii) ∫ \(\frac{\cos 2 x}{\sin ^2 x \cos ^2 x}\) dx
Solution:
(i) ∫ (√x – sin \(\frac{x}{2}\) cos \(\frac{x}{2}\) + 5) dx
= ∫ √x dx – ∫ \(\frac{x}{2}\) cos \(\frac{x}{2}\) dx + ∫ 5 dx
= \(\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\) – ∫ \(\frac{1}{2}\) sin x dx + 5x
[∵ sin 2θ = 2 sin θ cos θ]
= \(\frac{2}{3}\) x^3/2 + \(\frac{1}{2}\) cos x + 5x + C

(ii) ∫ \(\frac{\cos 2 x}{\sin ^2 x \cos ^2 x}\) dx
= \(\int \frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cos ^2 x}\) dx
= \(\int \frac{1}{\sin ^2 x} d x-\int \frac{1}{\cos ^2 x} d x\)
= ∫ cosec² x dx – ∫ sec² x dx
= – cot x – tan x + C

Question 16.
(i) ∫ \(\frac{\sec x}{\sec x+\tan x}\) dx
(ii) ∫ \(\frac{\ {cosec} x}{\ {cosec} x-\cot x}\) dx
Solution:
(i) ∫ \(\frac{\sec x}{\sec x+\tan x}\) dx
= ∫ \(\frac{\sec x(\sec x-\tan x)}{\sec ^2 x-\tan ^2 x}\) dx
= ∫ sec x (sec x – tan x) dx
[∵ sec² θ – tan² θ = 1]
= ∫ sec² x dx – ∫ sec x tan x dx + C
= tan x – sec x + C

(ii) ∫ \(\frac{\ {cosec} x}{\ {cosec} x-\cot x}\) dx
= ∫ \(\frac{\ {cosec} x(\ {cosec} x+\cot x) d x}{(\ {cosec} x-\cot x)(\ {cosec} x+\cot x)}\)
= ∫ \(\frac{\ {cosec}^2 x+\cot x \ {cosec} x d x}{\ {cosec}^2 x-\cot ^2 x}\)
[∵ 1 + cot² θ = cosec² θ ]
= ∫ cosec² x dx + ∫ cot x cosec x dx
= – cot x – cosec x + C

Question 17.
(i) ∫ \(\frac{x^6+1}{x^2+1}\) dx
(ii) ∫ \(\frac{6 \sin ^3 x+5 \cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
Solution:
(i) ∫ \(\frac{x^6+1}{x^2+1}\) dx
= ∫ \(\frac{\left(x^2+1\right)\left(x^4-x^2+1\right)}{x^2+1}\) dx
= ∫ (x⁴ – x² + 1) dx
= ∫ x⁴ dx + ∫ x² dx + ∫ 1 dx
= \(\frac{x^5}{5}-\frac{x^3}{3}\) + x + c
[∵ ∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + c ; n ≠ – 1]

(ii) ∫ \(\frac{6 \sin ^3 x+5 \cos ^3 x}{\sin ^2 x \cos ^2 x}\) dx
= 6 ∫ \(\frac{\sin ^3 x}{\sin ^2 x \cos ^2 x}\) dx + 5 ∫ \(\frac{\cos ^3 x d x}{\cos ^2 x \sin ^2 x}\)
= 6 ∫ tan x sec x dx + 5 ∫ cot x cosec x dx
= 6 sec x – 5 cosec x + C

Question 18.
(i) ∫ \(\sqrt{1+\cos 2 x}\) dx
(ii) ∫ \(\sqrt{1-\cos 2 x}\) dx
Solution:
(i) ∫ \(\sqrt{1+\cos 2 x}\) dx
= ∫ \(\sqrt{2 \cos ^2 x}\) dx
= √2 ∫ cos x dx
= √2 sin x + C

(ii) ∫ \(\sqrt{1-\cos 2 x}\) dx
= ∫ \(\sqrt{2 \sin ^2 x}\) dx
= √2 ∫ sin x dx
= – √2 cos x + C

Question 19.
(i) ∫ cot^-1 \(\left(\frac{\sin 2 x}{1-\cos 2 x}\right)\) dx
(ii) ∫ tan^-1 \(\left(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right)\) dx
Solution:
(i) ∫ cot^-1 \(\left(\frac{\sin 2 x}{1-\cos 2 x}\right)\) dx
= ∫ cot^-1 \(\left(\frac{2 \sin x \cos x}{2 \sin ^2 x}\right)\) dx
= ∫ cot^-1 (cot x) dx
= ∫ x dx
= \(\frac{x^{2}}{2}\) + C

(ii) ∫ tan^-1 \(\left(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right)\) dx
= ∫ tan^-1 \(\sqrt{\frac{2 \sin ^2 x}{2 \cos ^2 x}}\) dx
= ∫ tan^-1 (tan x) dx
= ∫ x dx
= \(\frac{x^{2}}{2}\) + C

Question 20.
(i) ∫ tan^-1 (cosec x – cot x) dx
(ii) ∫ cos^-1 \(\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right)\) dx
Solution:
(i) ∫ tan^-1 (cosec x – cot x) dx

(ii) ∫ cos^-1 \(\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right)\) dx
= ∫ cos^-1 (cos 2x) dx
[∵ cos 2θ = \(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\)]
= ∫ 2x dx
= x² + c

Question 21.
If f’(x) = 4x³ – 6 and f (0) = 3, find f (x).
Solution:
Given f’ (x) = 4x³ – 6
On integrating both sides w.r.t. x, we get
∫ f’(x) dx = ∫ (4x³ – 6) dx
f(x) = x⁴ – 6x + C ………..(1)
given f(0) = 3
i.e. when x = 0 ;
f(x) = 3
∴ from (1) ;
3 = c
Thus, from (1) ;
f(x) = x⁴ – 6x + 3.

Effective ISC Mathematics Class 12 Solutions Chapter 8 Integrals Ex 8.1 can help bridge the gap between theory and application.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.1

Very short answer type questions :

Evaluate the following integrals :

Question 1.
(i) ∫ x⁵ dx
(i) ∫ \(\frac{1}{x^3}\) dx
(iii) ∫ \(\frac{1}{\sqrt{y}}\) dy
Solution:
(i) ∫ x⁵ dx = \(\frac{x^{5+1}}{5+1}\) + C
[∵ ∫ x⁵ dx = \(\frac{x^{n+1}}{n+1}\) + C ; n ≠ 1]
= \(\frac{x^6}{6}\) + C

(ii) ∫ \(\frac{1}{x^3}\) dx
= ∫ x^{– 3} dx
= \(\frac{x^{-3+1}}{-3+1}\) + C
= \(\frac{x^{-2}}{-2}\) + C
= – \(\frac{1}{2 x^2}\) + C

(iii) ∫ \(\frac{1}{\sqrt{y}}\) dy
= ∫ y^{\(\frac{-1}{2}\)} dy
= \(\frac{y^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\) + C
= 2√y + C.

Question 2.
(i) ∫ \(\frac{1}{\sqrt[3]{x}}\) dx
(ii) ∫ sin² x cosec² x dx
(iii) ∫ sin x sec² x dx
Solution:
(i) ∫ \(\frac{1}{\sqrt[3]{x}}\) dx
= ∫ x^{– 1/3} dx
= \(\frac{x^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}\) + C
= \(\frac{3}{2}\) x^2/3 + C

(ii) ∫ sin² x cosec² x dx
= ∫ sin² x × \(\frac{1}{\sin ^2 x}\) dx
= ∫ 1 dx = x + C

(iii) ∫ sin x sec² x dx
= ∫ \(\frac{\sin x}{\cos ^2 x}\) dx
= ∫ tan x sec x dx
= sec x + C
[∵ ∫ tan x sec x dx = sec x + C]

Question 3.
(i) ∫ 5^x dx
(ii) ∫ 7^x e^x dx
(iii) ∫ \(\frac{4^x}{9^x}\) dx
Solution:
(i) ∫ 5^x dx
= \(\frac{5^x}{\log _e 5}\) + C
[∵ ∫ a^x dx
= \(\frac{a^x}{\log _e a}\) + C

(ii) ∫ 7^x e^x dx
= ∫ (7e)^x dx
= \(\frac{(7 e)^x}{\log 7 e}\) + C

(iii) ∫ \(\frac{4^x}{9^x}\) dx
= ∫ \(\left(\frac{4}{9}\right)^x\) dx
= \(\frac{\left(\frac{4}{9}\right)^x}{\log \frac{4}{9}}\) dx + C

Question 4.
(i) ∫ \(\frac{\sin x}{1-\sin ^2 x}\) dx
(ii) ∫ \(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\) dx (NCERT)
(iii) ∫ \(\sqrt{\frac{1}{2}(1+\cos 2 x)}\) dx
Solution:
(i) ∫ \(\frac{\sin x}{1-\sin ^2 x}\) dx
= ∫ \(\frac{\sin x}{\cos ^2 x}\) dx
= ∫ tan x sec x dx
= sec x + C

(ii) ∫ \(\frac{\cos 2 x+2 \sin ^2 x}{\cos ^2 x}\) dx
= ∫ \(\frac{1-2 \sin ^2 x+2 \sin ^2 x}{\cos ^2 x}\) dx
[∵ cos 2θ = 1 – 2 sin² θ]
= ∫ \(\frac{1}{\cos ^2 x}\) dx
= ∫ sec² x dx = tan x
[∵ ∫ sec² x dx = tan x + C]

(iii) ∫ \(\sqrt{\frac{1}{2}(1+\cos 2 x)}\) dx
= ∫ \(\sqrt{\frac{1}{2} \times 2 \cos ^2 x}\) dx
= ∫ cos x dx
= sin x + C

Question 5.
(i) ∫ e^{2 log x} x^{– 3} dx
(ii) ∫ 5^{2 log₅ x} dx
(iii) ∫ 2^{log₄ x} dx
Solution:
(i) ∫ e^{2 log x} x^{– 3} dx
= ∫ e^{log x2} x^{– 3} dx
= ∫ x² . x^{– 3} dx
= ∫ x^{– 1} dx
= ∫ \(\frac{1}{x}\) dx
= log |x| + C
[∵ ∫ \(\frac{1}{x}\) dx = log |x| + C]

(ii) ∫ 5^{2 log₅ x} dx
= ∫ 5^{log₅ x2} dx
= ∫ x² dx
= \(\frac{x^{3}}{3}\) + C

(iii) ∫ 2^{log₄ x} dx
= ∫ 2^{\(\frac{\log x}{2 \log 2}\)} dx
= ∫ \(2^{\frac{1}{2} \log _2 x}\) dx
= ∫ 2^{log₂ x1/2} dx
= ∫ x^{\(\frac{1}{2}\)} dx
= \(\frac{x^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right)}\) + C
= \(\frac{2}{3}\) x^{\(\frac{3}{2}\)} + C
[∵ ∫ xⁿ dx = \(\frac{x^{n+1}}{n+1}\) + C]

Students can cross-reference their work with Understanding ISC Mathematics Class 12 Solutions Chapter 7 Applications of Derivatives MCQs to ensure accuracy.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives MCQ’s

Choose the correct answer from the given four options in questions (1 to 47):

Question 1.
If the radius of a circle is increasing at the rate of 2 cm / sec, then the area of the circle when its radius is 20 cm is increasing at the rate of
(a) 80 π m² / sec
(b) 80 m² / sec
(c) 80 π cm² /sec
(d) 80 cm² /sec
Solution:
(c) 80 π cm² /sec

Let r be the radius of circle at any time t
Then A = area of circle = πr²
∴ \(\frac{d A}{d t}\) = 2 cm / sec
Given \(\frac{d r}{d t}\) = 2 cm/sec ;
r = 20 cm
∴ \(\frac{d A}{d t}\) = 2π × 20 × 2 = 80π cm² / sec

Question 2.
The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which its area increases, when side is 10 cm, is
(a) 10 cm² / sec
(b) 10√3 cm² /sec
(c) \(\frac{10}{3}\) cm² / sec
(d) √3 cm² / sec
Solution:
(b) 10√3 cm² /sec

Let x be the length of each side of equilateral triangle at any time t.
Then A = area of equilateral triangle = \(\frac{\sqrt{3}}{4}\) x²
∴ \(\frac{d \mathrm{~A}}{d t}=\frac{\sqrt{3}}{4} \times 2 x \frac{d x}{d t}\)
= \(\frac{\sqrt{3}}{2} x \frac{d x}{d t}\)
Given \(\frac{d x}{d t}\) = 2 cm / sec ;
x = 10 cm
∴ \(\frac{d A}{d t}\) = \(\frac{\sqrt{3}}{2}\) × 10 × 2 = 10√3 cm² / sec.

Question 3.
A spherical ice ball is melting at the rate of 100 π cm³ / min. The rate at which its radius is decreasing, when its radius is 15 cm, is
(a) \(\frac{1}{9}\) cm / min
(b) \(\frac{1}{9 \ pi}\) cm / min
(c) \(\frac{1}{18}\) cm / min
(d) \(\frac{1}{36}\) cm / min
Solution:
(a) \(\frac{1}{9}\) cm / min

Let r be the radius of spherical ice ball
Then \(\frac{d V}{d t}\) = – 100π cm³ / min.
⇒ \(\frac{d}{d t}\) (\(\frac{4}{3}\) πr³) = – 100π
⇒ \(\frac{4}{3}\) π × 3r² \(\frac{d r}{d t}\) = – 100π
⇒ 4π × 15² . \(\frac{d r}{d t}\) = – 100π [given r = 15]
⇒ \(\frac{d r}{d t}\) = – \(\frac{1}{9}\) cm / min.

Question 4.
The radius of a cylinder is increasing at the rate of 3 cm/sec and its height is decreasing at the rate of 4 cm/sec. The rate of change of its volume when radius is 4 cm and height is 6 cm, is
(a) 64 cm³ / sec
(b) 144 cm³/ sec
(c) 80 cm³/ sec
(d) 80 π cm³ / sec
Solution:
(d) 80 π cm³ / sec

Let r be the radius and h be the height of cylinder at any time.
Then V = volume of cylinder = πr²h
∴ \(\frac{d V}{d t}=\pi\left[r^2 \frac{d h}{d t}+2 r h \frac{d r}{d t}\right]\)
Given \(\frac{d r}{d t}\) = 3 cm/sec ;
\(\frac{d h}{d t}\) = – 4 cm/sec ;
r = 4 cm
and h = 6 cm
Thus \(\frac{d V}{d t}\) = π [4² × (- 4) + 2 × 4 × 6 × 3]
= π [- 64 + 144]
= 80 π cm³ / sec

Question 5.
A point on the curve y² = 18x at which the ordinate increases at twice the rate of abscissa is
(a) (2, 4)
(b) (2, – 4)
(c) \(\left(-\frac{9}{8}, \frac{9}{2}\right)\)
(d) \(\left(\frac{9}{8}, \frac{9}{2}\right)\)
Solution:
(d) \(\left(\frac{9}{8}, \frac{9}{2}\right)\)

Given eqn. of curve y² = 18x
⇒ 2y \(\frac{d y}{d x}\) = 18
⇒ \(\frac{d y}{d x}=\frac{9}{y}\)
⇒ 2y \(\frac{d y}{d t}\) = 18 \(\frac{d x}{d t}\)
⇒ \(\frac{d y}{d t}=\frac{9}{y} \frac{d x}{d t}\) ……..(1)
Also \(\frac{d y}{d t}\) = 2 \(\frac{d x}{d t}\)
from (1) and (2) ; we have
⇒ 2 \(\frac{d x}{d t}\) = \(\frac{9}{y}\) \(\frac{d x}{d t}\)
⇒ y = \(\frac{9}{2}\) \(
∴ from given curve, we have,
[latex]\left(\frac{9}{2}\right)^2\) = 18x
⇒ x = \(\frac{81}{4 \times 18}=\frac{9}{8}\)
Thus required point on given curve be \(\left(\frac{9}{8}, \frac{9}{2}\right)\).

Question 6.
A ladder, 5 metres long standing on a floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when the lower end of the ladder is 2 metres away from the wall is
(a) \(\frac{1}{10}\) radian / sec
(b) \(\frac{1}{20}\) radian/sec
(c) 5 radian/sec
(d) 10 radian/sec
Solution:
(b) \(\frac{1}{20}\) radian/sec

Let the bottom of the ladder be at a distance x m from the wall and the top be at a height h from the ground.
Then x² + y² = 25 ………..(1)
and tan θ = \(\frac{y}{x}\) …………(2)
Diff. eqn. (1) both sides w.r.t. t, we have
2x \(\frac{d x}{d t}\) + 2y \(\frac{d y}{d t}\) = 0
Given \(\frac{d y}{d t}\) = – 10 cm / sec = – \(\frac{1}{10}\) m / sec
⇒ 2x \(\frac{d x}{d t}\) – \(\frac{2 y}{10}\) = 0
⇒ \(\frac{d x}{d t}=\frac{y}{10 x}\)
Diff. eqn. (2) both sides w.r.t. t, we have

Question 7.
The curve y = x^1/5 has at (0, 0)
(a) a vertical tangent (parallel to y-axis)
(b) a horizonlal tangent (parallel to x-axis)
(c) an oblique tangent
(d) no tangent
Solution:
(a) a vertical tangent (parallel to y-axis)

Equation of given curve be,
y = x^1/5
∴ slope of tangent to given curve at point
P(0, 0) = (\(\frac{d y}{d x}\))_{(0, 0)} → ∞
Thus tangent is parallel to y-axis.

Question 8.
The equation of the tangent to the curve y = (4 – x²)^2/3 at x = 2 is
(a) x = – 2
(b) x = 2
(c) y = 2
(d) y = – 2
Solution:
(b) x = 2

eqn. of given curve be y = (4 – x²)^2/3
When x = 2 ; y = 0
Thus any point on given curve be (2, 0)
∴ \(\frac{d y}{d x}\) = \(\frac{2}{3}\) (4 – x²)^-1/3 (- 2x)
∴ slope of tangent to given curve at x = 2 = (\(\frac{d y}{d x}\))_{x = 2} = ∞
Thus required eqn. of tangent to given curve at (2, 0) be given by
y – 0 = \(\frac{\frac{1}{d x}}{d y}\) (x – 2)
⇒ x – 2 = 0
⇒ x = 2.

Question 9.
The equation of the tangent to the curve y = e^2x at (0, 1) is
(a) y + 1 = 2x
(b) 1 – y = 2x
(c) y – 1 = 2x
(d) none of these
Solution:
(c) y – 1 = 2x

eqn. of given curve be
y = e^2x …………(1)
∴ \(\frac{d y}{d x}\) = 2e^2x
Thus slope of tangent to curve (1) at (0, 1) = (\(\frac{d y}{d x}\))_{(0, 1)}
=2 e^{2 × 0} = 2
Thus eqn. of tangent to given curve at point (0, 1) be given by
y – 1 = 2(x – 0)
⇒ y – 1 = 2x.

Question 10.
The tangent to the curve y = e at the point (0, 1) meets x-axis at:
(a) (0, 1)
(b) (- \(\frac{1}{2}\), o)
(c) (2, 0)
(d) (0, 2)
Solution:
(b) (- \(\frac{1}{2}\), o)

Equation of given curve be y = e^2x ………….(1)
Differentiate equation (1) both sides w.r.t. x; we get
\(\frac{d y}{d x}\) = 2e^2x
∴ slope of tangent to given curve (1) at (0, 1) = (\(\frac{d y}{d x}\))_{(0, 1)}
= 2 × e⁰ = 2
Thus required equation of tangent to given curve (1) at (0, 1) is given by
y – 1 = 2 (x – 0)
⇒ y – 1 = 2x …………(2)
Equation (2) meets x-axis at y = 0
Hence the coordinates of required point are (- \(\frac{1}{2}\), o).

Question 11.
The tangent to the curve x² = 2y at the point (1, \(\frac{1}{2}\)) makes with the x-axis an angle of
(a) 0
(b) \(\frac{\pi}{6}\)
(c) \(\frac{\pi}{4}\)
(d) \(\frac{\pi}{3}\)
Solution:
(c) \(\frac{\pi}{4}\)

Given eqn. of curve be x² = 2y
∴ 2x = 2 \(\frac{d y}{d x}\)
⇒ \(\frac{d y}{d x}\) = x
∴ slope of tangent to given curve at (1, \(\frac{1}{2}\)) = (\(\frac{d}{d x}\))_{(1, \(\frac{1}{2}\))} = 1
Let θ be the angle made by tangent with the +ve direction of x-axis.
∴ slope of tangent to given curve = tan θ
∴ tan θ = 1
⇒ θ = \(\frac{\pi}{4}\)

Question 12.
The tangents to the curve x² + y² = 2 at points (1, 1) and (- 1, 1) are
(a) parallel
(b) at right angles
(c) neither parallel nor at right angles
(d) none of these
Solution:
eqn. of given curve be
x² + y² = 2 …………(1)
Diff. eqn. (1) w.r.t. x; we have
2x + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = – \(\frac{x}{y}\)
∴ m₁ = slope of tangent to given curve at (1, 1) = (\(\frac{d y}{d x}\))_{(0, 1)} = – 1
m₂ = slope of tangent to given curve at (- 1, 1) = (\(\frac{d y}{d x}\))_{(- 1, 1)} = 1
∴ m₁ m₂ = – 1 × 1 = – 1
Hence, tangents to given curve at points (1, 1) and (- 1, 1) are at right angles.

Question 13.
The point on the curve y² = x, where tangent make an angle of with the x-axis, is
(a) \(\left(\frac{1}{2}, \frac{1}{4}\right)\)
(b) \(\left(\frac{1}{4}, \frac{1}{2}\right)\)
(c) (4, 2)
(d) (1, 1)
Solution:
(b) \(\left(\frac{1}{4}, \frac{1}{2}\right)\)

eqn. of given curve be
y²= x …………….(1)
diff, both sides of eqn. (1) w.r.t. x;
2y \(\frac{d y}{d x}\) = 1
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2 y}\)
∴ slope oftangentto curve (1) = \(\frac{d y}{d x}\) = \(\frac{1}{2 y}\)
Also, slope of tangent to given curve (1) = tan \(\frac{\pi}{4}\) = 1
∴ \(\frac{1}{2 y}\) = 1
⇒ y = \(\frac{1}{2}\)
∴ from (1) ;
x = \(\frac{1}{4}\)
Thus, required point on given curve be (\(\left(\frac{1}{4}, \frac{1}{2}\right)\))

Question 14.
The point on the curve y = 6x – x² where the tangent is parallel to the line 4x – 2y – 1 = 0 is
(a) (2, 8)
(b) (8, 2)
(c) (6, 1)
(d) (4, 2)
Solution:
(a) (2, 8)

Let P (x₁, y₁) be any point on curve
y = 6x – x² ……….(1)
∴ \(\frac{d y}{d x}\) = 6 – 2x
Thus slope of tangent to given curve at
P(x₁, y₁) = \(\left(\frac{d y}{d x}\right)_{\mathrm{P}\left(x_1, y_1\right)}\) = 6 – 2x₁
eqn.of given line be 4x – 2y – 1 = 0 ………….(2)
∴ slope of hne (2) = \(\frac{- 4}{- 2}\) = 2
Since it is given that tangent to given curve is parallel to line (2)
∴ 6 – 2x₁ = 2
⇒ 2x₁ = 4
⇒ x₁ = 2
Also P (x₁, y₁) lies on eqn. (1) ;
y₁ = 6x₁ – x₁
= 6 × 2 + 2² = 8
Hence coordinates of any point on given curve be (2, 8).

Question 15.
The points at which the tangents to the curve y = x³ – 12x + 18 are parallel to x-axis are :
(a) (2, – 2), (- 2, – 34)
(b) (2, 34), (- 2, 0)
(c) (0, 34), (- 2, 0)
(d) (2, 2), (- 2, 34)
Solution:
(d) (2, 2), (- 2, 34)

Equation of given curve be
y = x³ – 12x + 18 …………..(1)
Let the required points on curve (1) are (x₁, y₁)
Differentiate equation (1) w.r.t. x; we get
\(\frac{d y}{d x}\) = 3x² – 12
slope of tangent to given curve (1) at (x₁, y₁)
= \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= 3x₁² – 12
Also tangents are parallel to x – axis.
∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\) = 0
[since slope of the tangent is 0]
3x₁² – 12 = 0
⇒ x₁ = ± 2
Also, the point (x₁, y₁) lies on curve (1)
y₁= x₁ – 12x₁ + 18 …………….(2)
when x₁ = 2
∴ from (2) ; we have
y₁ = 8 – 24 + 18 = 2
when x₁ = 2
∴ from (2) ; we have
y₁ = – 8 + 24 + 18 = 34
Thus the required points on given curve (1) are (2, 2) and (- 2, 34).

Question 16.
The point at which the tangent to the curve y = 2x² – x + 1 is parallel to the line y = 3x + 9 is
(a) (2, 1)
(b) (1, 2)
(c) (3, 9)
(d) (- 2, 1)
Solution:
(b) (1, 2)

Let P (x₁, y₁) be any point on given curve
y = 2x² – x + 1
∴ y₁ = 2x₁² – x₁ + 1 ………..(1)
∴ slope of tangent to given curve at P (x₁, y₁)
= \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)
= 4x₁ – 1
Again eqn. of given line be 3x – y + 9 = 0 …………..(2)
∴ slope of line (2) = \(\frac{- 3}{- 1}\) = 3
Since tangent to curve (1) is parallel to line (2)
∴ 4x₁ – 1 = 3
⇒ x₁ = 1
∴ from (1);
y₁ = 2 – 1 + 1 = 2
Thus, required point on tangent to given curve be (1, 1).

Question 17.
If the tangent to the curve x = t² – 1, y = t² – t is parallel to x-axis, then
(a) t = 0
(b) t = 2
(c) t = \(\frac{1}{2}\)
(d) t = – \(\frac{1}{2}\)
Solution:
(c) t = \(\frac{1}{2}\)

Given eqn. of curve be,
x = t² – 1
and y = t² – t
∴ \(\frac{d x}{d t}\) = 2t ;
\(\frac{d y}{d t}\) = 2t – 1
Thus \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{2 t-1}{2 t}\)
Since tangent to given curve is parallel to x-axis
∴ \(\frac{d y}{d x}\) = 0
⇒ \(\frac{2 t-1}{2 t}\) = 0
⇒ t = \(\frac{1}{2}\).

Question 18.
The tangent to the curve x = e^t cos t, y = e^t sin t at t = \(\frac{\pi}{4}\) makes with x-axis an angle
(a) 0
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{\pi}{2}\)
Solution:
(d) \(\frac{\pi}{2}\)

Given parametric eqns. of curve be,
x = e^t cos t ……………..(1)
y = e^t sin t …………..(2)
Diff. eqns. (1) and (2) w.r.t. t; we have
\(\frac{d x}{d t}\) = e^t (- sin t) + cos t . e^t
and \(\frac{d y}{d t}\) = e^t cos t + sin t e^t
∴ \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{e^t(\cos t+\sin t)}{e^t(\cos t-\sin t)}\)
= \(\frac{\cos t+\sin t}{\cos t-\sin t}\)
∴ slope of tangent to given curve at t = \(\frac{\pi}{4}\)
= \(\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{4}}\)
= \(\frac{\cos \frac{\pi}{4}+\sin \frac{\pi}{4}}{\cos \frac{\pi}{4}-\sin \frac{\pi}{4}}\)
= \(\frac{\sqrt{2}}{0}\) → ∞
Let the tangent to given curve at t = \(\frac{\pi}{4}\) makes an angle θ with x – axis.
Then slope of tangent to given curve at t = \(\frac{\pi}{4}\) = tan θ
∴ tan θ = ∞
⇒ θ = \(\frac{\pi}{2}\).

Question 19.
The equation to the normal to the curve sin x at (0, 0) is
(a) x = 0
(b) y =
(c) x + y = 0
(d) x – y = 0
Solution:
(c) x + y = 0

Given eqn. of curve be y = sin x …………..(1)
∴ \(\frac{d y}{d x}\) = cos x
∴ \(\left(\frac{d y}{d x}\right)_{(0,0)}\) = cos 0 = 1
Thus required equation of normal at (0, 0) to given curve be
y – 0 = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{(0,0)}}\) (x – 0)
⇒ y = – 1 (x)
⇒ x + y = 0.

Question 20.
The slope of the normal to the curve X2 + 3y +ý = 5 at the poInt (1, 1) ¡s
(a) – \(\frac{2}{5}\)
(b) \(\frac{5}{2}\)
(c) \(\frac{2}{5}\)
(d) – \(\frac{5}{2}\)
Solution:
(b) \(\frac{5}{2}\)

eqn. of given curve be
x² + 3y + y² = 5 …………(1)
duff. eqn. (1) both sides w.r.t. X; we have
2x + 3 \(\frac{d y}{d x}\) + 2y \(\frac{d y}{d x}\) = 0
⇒ (2y + 3) \(\frac{d y}{d x}\) = – 2x
⇒ \(\frac{d y}{d x}\) = \(\frac{-2 x}{2 y+3}\)
∴ slope of normal to given curve at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{(1,1)}}\)
= \(\frac{-1}{\frac{-2 \times 1}{2 \times 1+3}}\)
= \(\frac{5}{2}\).

Question 21.
The point on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes is
(a) (4, ± 8/3)
(b) (- 4, 8/3)
(c) (- 4, – 8/3)
(d) (8/3, 4)
Solution:
(a) (4, ± 8/3)

Given eqn. of curve be
9y² = x³ ………….(1)
Let the required point on curve (1) be (x₁, y₁)
∴ 9y₁² = x₁³ ………….(2)
Diff. eqn. (1) both sides w.r.t. x; we get
18 y = \(\frac{d y}{d x}\) = 3x²
Since normal to curve makes an equal intercept with axes.
∴ slope of normal at (x₁, y₁) = ± 1
∴ \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}}\) = ± 1
⇒ \(\frac{-6 y_1}{x_1^2}\) = ± 1
⇒ \(\frac{6 y_1}{x_1^2}\) = ± 1
when \(\frac{6 y_1}{x_1^2}\) = 1
⇒ 6y₁ = x₁² ………….(3)
∴ from (2) ; we have
\(\frac{9 x_1^4}{36}\) = x₁³
⇒ x₁³ (x₁ – 4) = 0
⇒ x₁ = 0, 4
When x₁ = 0
∴ from (3) ;
y₁ = 0
When x₁ = 4
∴ from (2) ;
y₁ = ± 8/3
When \(\frac{6 y_1}{x_1^2}\) = – 1
⇒ 6y₁ = – x₁² …………(4)
∴ from (2) ;
x₁ = 0, 4
When x₁ = 4
∴ from (4) ;
y₁ = 0
When x₁ = 0
∴ from (4) ; y₁ = 0
Thus required points are (0. 0), (4, ± 8/3).

Question 22.
The equation of the normal to the curve 3x² – y² = 8 which is parallel to x + 3y = 8 is
(a) x – 3y = 8
(b) x – 3y + 8 = 0
(c) x + 3y ± 8 = 0
(d) x + 3y = 0
Solution:
Given eqn. of curve be 3x² – y² = 8 ………..(1)
Let (x₁, y₁) be point on curve (1) where normal to curve (1) || to given line
x+3y – 8 = 0
Diff. (1) w.r.t. x; we get
6x – 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{d3 x}{y}\)
∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=\frac{3 x_1}{y_1}\)
Since the normal to curve (1) || to line x + 3y – 8 = 0
∴ slope of normal at (x₁, y₁) = slope of line
x + 3y – 8 = 0
⇒ \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}}=-\frac{1}{3}\)
⇒ \(-\frac{y_1}{3 x_1}=-\frac{1}{3}\)
⇒ x₁ = y₁ …………….(2)
also point (x₁, y₁) lies on eqn. (1)
∴ 3x₁² – y₁² = 8
⇒ 3x₁² – x₁² = 8
⇒ x₁² = 4
⇒ x₁ = ± 2
from (2) ;
y₁ = ± 2
Thus point of contact be (± 2, ± 2).
∴ required eqn. of normal at (2, 2) to given curve be
y – 2 = – (x – 2)
3y – 6 = – x + 2
x + 3y -8 = 0
and eqn. of normal at (- 2, – 2) to given curve be
y – 2 = – \(\frac{1}{3}\) (x – 2)
⇒ 3y – 6 = – x + 2
⇒ x + 3y – 8 = 0
and eqn. of normal at (- 2, – 2) to given curve be
y + 2 = – \(\frac{1}{3}\) (x + 2)
⇒ x + 3y + 8 = 0

Question 23.
The angle between the tangents to the curve y = x² – 5x + 6 at the point (2, 0) and (3, 0) is
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{4}\)
(d) \(\frac{\pi}{6}\)
Solution:
(a) \(\frac{\pi}{2}\)

eqn. of given curve be y = x² – 5x + 6
∴ \(\frac{d y}{d x}\) = 2x – 5
m₁ = slope of tangent to given curve at (2, 0)
= \(\left(\frac{d y}{d x}\right)_{(2,0)}\)
= 2 × 2 – 5 = – 1
and m₂ = slope of tangent to given curve at (3, 0)
= \(\left(\frac{d y}{d x}\right)_{(2,0)}\)
= 2 × 3 – 5 = 1
Here m₁m₂ = (- 1 ) × 1 = – 1
Thus both tangents to given curve at points (2, 0) and (3, 0) cut orthogonally
i.e. the angle between them is \(\frac{\pi}{2}\).

Question 24.
If the curve ay + x² = 7 and x³ = y cut orthogonally at (1, 1), then a is equal to
(a) 1
(b) – 6
(c) 6
(d) 0
Solution:
(c) 6

Given eqns. of curve be,
ay + x² = 7 ……….(1)
and x³ = y ………..(2)
Diff. (1) and (2) both sides w.r.t. x, we get
a \(\frac{d y}{d x}\) + 2x = 0
∴ \(\frac{d y}{d x}\) = – \(\frac{2 x}{a}\)
∴ m₁ = \(\left(\frac{d y}{d x}\right)_{(1,1)}\)
= \(-\frac{2}{a}\)
and 3x² = \(\frac{d y}{d x}\)
∴ m₂ = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) = 3
Since both cuves cut orthogonally.
∴ m₁m₂ = – 1
⇒ – \(\frac{2}{3}\) × a = – 1
⇒ a = 6

Question 25.
The two curves x³ – 3xy² + 2 = 0 and 3x²y – y³ – 2 = 0 intersect at an angle of:
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{2}\)
(d) \(\frac{\pi}{6}\)
Solution:
(c) \(\frac{\pi}{2}\)

Given eqns. of curves be,
x³ – 3xy² = – 2 ……………(1)
and 3x²y – y³ = 2 …………..(2)
From (1) and (2) ; we get
x³ – 3xy² = – 3x²y + y³
⇒ x³ – 3xy² + 3x²y – y³ = 0
⇒ (x – y)³ = 0
⇒ x = y
putting x =y in eqn. (1); we get
x³ – 3x³ = – 2
⇒ x³ = 1
⇒ x = + 1
∴ from (3) ;
y = + 1
Hence the points of intersection of two curves be (1, 1).
Diff. (1) w.r.t. x; we get
3x² – 3 [2xy \(\frac{d y}{d x}\) + y²] = 0
⇒ \(\frac{d y}{d x}=\frac{x^2-y^2}{2 x y}\) ……..(3)
Diff. (2) w.r.t. x; we get
3 [x² \(\frac{d y}{d x}\) + y . 2x] – 3y² \(\frac{d y}{d x}\) = 0
\(\frac{d y}{d x}=-\frac{2 x y}{x^2-y^2}\) ………….(4)

at (1, 1) :
using (3);
m₁ = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) = 0
m₂ = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) = ∞
If θ be the angle of intersection between two curves.
Then tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
= \(\left|\frac{m_2\left(\frac{m_1}{m_2}-1\right)}{m_2\left(\frac{1}{m_2}+m_1\right)}\right|\) = ∞
⇒ θ = \(\frac{\pi}{2}\)
Hence both curves intersect orthogonally at (1, 1).

Question 26.
The interval on which the function f(x) = 2x³ + 9x² + 12x – 1 is decreasing in:
(a) [- 1, ∞)
(b) [- 2, – 1]
(c) (- ∞, – 2]
(d) [- 1, 1]
Solution:
(b) [- 2, – 1]

Given f (x) = 2x³ + 9x² + 12x + 20
∴ f’(x) = 6x² + 18x + 12
= 6 (x² + 3x + 2)
= 6 (x + 1) (x + 2)
Now f’(x) > 0
⇒ (x + 1) (x + 2) > 0
⇒ x > – 1 or x < – 2
∴ f(x) is strictly increasing in (- ∞, – 2) ∪ (- 1, ∞).
Now f’(x) < 0
⇒ (x + 1) (x + 2) < 0
⇒ – 2 < x < – 1
[If (x – a) (x – b) < 0 and a < b ⇒ a < x < b]
∴ f(x) is strictly decreasing in [- 2, – 1].
[If f’(x) < 0 ∀ x ∈ (a, b)Thenf(x) is decreasg in [a, b]]

Question 27.
The interval in which the function f(x) = 2x³ + 3x² – 12x + 1 is strictly increasing is
(a) [- 2, 1]
(b) (- ∞, – 2] ∪ [1, ∞)
(e) (- ∞, 1]
(f)(- ∞, – 1] ∪ [2, ∞)
Solution:
(b) (- ∞, – 2] ∪ [1, ∞)

Given f(x) = 2x³ + 3x² – 12x + 1
∴ f’(x) = 6x² + 6x – 12
= 6 (x² + x – 2)
= 6 (x – 1) (x + 2)
Now f’(x) > 0 iff 6 (x – 1) (x + 2) > 0
iff x > 1 or x< – 2 [if (x – a) (x – b) > 0 and a > b ⇒ x > a or x < b]
iff x ∈ (- ∞, – 2) ∪ (1, ∞)
Thus f(x) is strictly increasing in (- ∞, – 2] ∪ [1, x).

Question 28.
The function f (x) = x² e^{– x} is monotonic increasing when
(a) x ∈ R – [0, 2]
(b) 0 < x < 2
(c) 2 < x < ∞
(d) x < 0
Solution:
(b) 0 < x < 2

Given f(x) = x² e^{– x}
∴ f'(x) = x² e^{– x} (- 1) + e^{– x} 2x
= e^{– x} [- x² + 2x]
= e^{– x} (2 – x) x
For f(x) is to be monotomically decreasing we must have
f’(x) ≥ 0
⇒ e^{– x} (2 – x) x ≥ 0 [where e^{– x} ≥ 0]
⇒ – (x – 2)x ≤ 0
⇒ (x – 2) x ≤ 0
⇒ 0 < x < 2
⇒ x ∈ (0, 2)
[if (x – a) (x – b)< 0 and a < b then a < x < b].

Question 29.
The function f(x) = tan x – x
(a) always increases
(b) always decreases
(c) never increases
(d) sometimes increases and decreases.
Solution:
(a) always increases

Given f(x) = tan x – x
∴ f'(x) = sec² x – 1 ≥ 0
[sec² x ≥ 1
⇒ sec² x – 1 ≥ 0 ∀ x ∈ R]
Hence the function f(x) is always increases.

Question 30.
The function f(x) = x⁴ – 4x is strictly
(a) decreasing in [1, ∞)
(b) increasing in [1, ∞)
(c) Increasing in (- ∞, 1]
(d) increasing in [- 1, 1]
Solution:

Given f(x) = x⁴ – 4x
f’(x) = 4x³ – 4
= 4 (x³ – 1)
= 4 (x – 1) (x² + x + 1)
= 4 (x – 1) (x² + x + \(\frac{1}{4}\) + \(\frac{3}{4}\))
= 4 (x – 1) [(x + \(\frac{1}{2}\))² + \(\frac{3}{4}\)]
Now f’ (x) > 0
iff 4 (x – 1) [(x + \(\frac{1}{2}\))² + \(\frac{3}{4}\)] > 0
iff x – 1 > 0
[∵ (x + \(\frac{1}{2}\)) + \(\frac{3}{4}\) > 0 ∀ x ∈ R]
iff x > 1
iff x ∈ (1, ∞)
Thus f(x) is strictly increasing in [1, ∞)

Question 31.
The function f (x) = x² – 2x is strictly decreasing in the interval
(a) (- ∞, 1]
(b) [1, ∞)
(c) [- 1, ∞)
(d) none of these
Solution:
(a) (- ∞, 1]

Given f(x) = x² – 2x
∴ f’(x) = 2x – 2
Now f’(x) < 0 iff 2x – 2 < 0
iff x < 1 iff x ∈ (- ∞, 1)
Thus,f(x) is strictly decreasing in (- ∞, 1].

Question 32.
y = x (x – 3)² decreases for the values of x given by:
(a) 1 < x < 3
(b) x < 0 (c) x > 0
(d) 0 < x < \(\frac{3}{2}\)
Solution:
(a) 1 < x < 3

Given y = x (x – 3)² = f(x)
∴ \(\frac{d y}{d x}\) = (x – 3)² + 2x (x – 3)
= (x – 3) (x – 3 + 2x)
= 3 (x – 3) (x – 1)
since f(x) is decreases if f’ (x) < 0
i.e. \(\frac{d y}{d x}\) < 0
⇒ (x – 3) (x – 1) < 0
⇒ 1 < x < 3
[if a < b and (x – a) (x- b) < 0 ⇒ a < x < b]

Question 33.
Which of the following functions is decreasing on (o, \(\frac{\pi}{2}\))
(a) sin 2x
(b) tan x
(c) cos x
(d) cos 3x
Solution:
When 0 < x < \(\frac{\pi}{2}\)
⇒ 0 < 2x < π For option (A) ; f (x) = sin 2x ∴ f’(x) = 2 cos 2x may be > 0 or < 0.
Hence f(x) may be increasing or decreasing.
For option (B) ;
f (x) = tan x
∴ f'(x) = sec² x > 0 ∀ x ∈ (0)
f(x) is increasing on (o, \(\frac{\pi}{2}\)).
For option (C) ;
f (x) = cos x
∴ f’(x) = – sin x < 0 ∀ x ∈ (o, \(\frac{\pi}{2}\))
∴ f(x) is decreasing on (o, \(\frac{\pi}{2}\)).
For option (D) ;
f (x) = cos 3x
∴ f ‘(x) = 3 sin 3x
[∵ 0 < x < \(\frac{\pi}{2}\) ⇒ 0 < 3x < \(\frac{3 \pi}{2}\)]
it may be positive or negative.
Thus f(x) may be increasing or decreasing.

Question 34.
The function f (x) = x^x, x > 0, is increasing on the interval
(a) (0, e]
(b) (0, \(\frac{1}{e}\))
(c) [\(\frac{1}{e}\), ∞)
(d) none of these
Solution:
(c) [\(\frac{1}{e}\), ∞)

Given f(x) = x^x
∴ log f(x) = Iog x^x = x log x;
Diff, both sides w.r.t. x, we have
⇒ \(\frac{1}{f(x)}\) f’(x) = x × \(\frac{1}{x}\) + log x × 1
⇒ f’ (x) = x^x (1 + log x)
Now f’(x) ≥ 0 iff x^x (1 + log x) ≥ 0
iff 1 + log x ≥ 0 [∵ x^x > 0]
iff log x ≥ 1 iff x ≥ \(\frac{1}{e}\)
Thus f(x) is increasing in [\(\frac{1}{e}\), ∞).

Question 35.
The value of p so that the function f (x) = sin x – cos x – px + q decreases for all real values of x is given by
(a) p ≥ √2
(b) p < √2
(c) p ≥ 1
(d) p < 1
Solution:
(a) p ≥ √2

f’(x) = cos x + sin x – p
f(x) is decreasing iff f’cos ≤ 0
iff cosx + sinx – p ≤ 0
iff cos x + sin x ≤ p
iff √2 sin (x + \(\frac{\pi}{4}\)) ≤ p
Since |sin x| ≤ 1
∴ p ≥ √2

Question 36.
If x is real, x² – 8x + 17 is
(a) – 1
(b) 0
(c) 1
(d) 2
Solution:
(c) 1

Let f(x) = x² – 8x + 17
= x² – 8x + 16 + 1
= (x – 4)² + 1 ≥ 1 ∀ x ∈ R
Hence 1 be the minimum value of f (x) and is attains at x = 4.

Question 37.
The smallest value of the polynomial x³ – 18x² + 96x in [0, 9] is
(a) 126
(b) 135
(c) 160
(d) 0
Solution:
(d) 0

Given f(x) = x³ – 18x² + 96x
∴ f’(x) = 3x² – 36x+96
= 3 (x² – 12x + 32)
= 3 (x – 8) (x – 4)
For critical points, f’ (x) = 0
⇒ 3 (x – 8) (x – 4) = 0
⇒ x = 4, 8
Now we compute f (x) at critical points x = 4, 8 and at end points x = 0, 9.
Here, f(0) = 0;
f(4) = 4³ – 18 x 4² + 96 x 4
= 64 – 288 + 384
= 160
f(8) = 8³ – 18 × 64 + 96 × 8 = 128
f(9) = 9³ – 18 × 81 + 96 × 9 = 135
∴ smallest value of f(x) = f(0) = 0.

Question 38.
The minimum value of (x² + \(\frac{250}{x}\)) is
(a) 75
(b) 50
(c) 25
(d) 55
Solution:
(a) 75

Let f(x) = x² + \(\frac{250}{x}\)
∴ f’(x) = 2x – \(\frac{250}{x^{2}}\)
For maxima/minima, f’ (x) = 0
⇒ 2x – \(\frac{250}{x^{2}}\) = 0
⇒ x³ = 125 = 5³
[∵ x³ – 5³
⇒ (x – 5) (x² + 5x + 25) = 0
⇒ x = 5 and x² + 5x + 25 = 0 does not gives any real values of x]
⇒ f”(x) = 2 + \(\frac{500}{x^{3}}\)
∴ f”(5) = 2 + \(\frac{500}{125}\)
= 6 > 0
Hence x = 5 is a point of minima and min.
value of f(x) = f(5) = 25 + \(\frac{250}{5}\) = 75

Question 39.
The function f(x) = \(\frac{x}{2}+\frac{2}{x}\) has a local minimum at
(a) x = 1
(b) x = 2
(c) x = – 2
(d) x = – 1
Solution:
(b) x = 2

Given f(x) = \(\frac{x}{2}+\frac{2}{x}\)
∴ f'(x) = \(\frac{1}{2}-\frac{2}{x^2}\)
For local maxima/minima, f'(x) = 0
⇒ \(\frac{1}{2}-\frac{2}{x^2}\) = 0
⇒ \(\frac{1}{2}=\frac{2}{x^2}\)
⇒ x² = 4
⇒ x = ± 2
at x = 2;
f”(2) = \(\frac{4}{2^3}=\frac{1}{2}\) >0
Thus f(x) is minimise when x = 2.

Question 40.
Let the function f: R → R be defined by f(x) = 2 x+ cos x, then f(x)
(a) has a minimum at x = it
(b) has a maximum at x = 0
(c) is a decreasing function
(d) is an increasing function [NCERT Exemplar]
Solution:
(d) is an increasing function

Given f(x) = 2x + cos x
∴ f’(x) = 2 – sinx
Now – 1 ≤ sin x ≤ 1
⇒ 1 ≥ – sin x ≥ – 1
⇒ 3 ≥ 2 – sin x ≥ 1
⇒ 1 ≤ f’(x) ≤ 3
∴ f'(x) > 0
Thus f(x) is an increasing function.

Question 41.
At x = \(\frac{5 \pi}{6}\), f(x) = 2 sin 3x + 3 cos 3x is
(a) 0
(b) maximum
(c) minimum
(d) none of these
Solution:
(d) none of these

f(x) = 2 sin 3x + 3 cos 3x
f’(x) = 6 cos 3x – 9 sin 3x
f” (x) = – 18 sin 3x – 27 cos 3x
∴ f’(5 π / 6) = 6 cos \(\frac{5 \pi}{2}\) – 9 sin 5π /2
= 6 cos (2π+ π/2) – 9 sin (2π + π/2)
= 6 × 0 – 9 × 1
= – 9 ≠ 0
f(5 π / 6) = 2 sin 5 π / 2 + 3 cos 5 π / 2
= 2 × 1 + 3 × 0
=2 ≠ 0
Thus x = \(\frac{5 \pi}{2}\) is not an extreme point.

Question 42.
The f(x) = x^x has a stationary point at
(a) x = e
(b) x = \(\frac{1}{e}\)
(c) x = 1
(d) x = √e
Solution:
(b) x = \(\frac{1}{e}\)

Given f (x) = x^x;
Taking logarithm on both sides ; we have
∴ log f(x) = x log x
Differentiate both sides wr.t. x; we have
\(\frac{1}{f(x)}\) f’(x) = x . \(\frac{1}{x}\) + log x . 1
⇒ f ‘(x) = f(x) [1 + log x]
= x^x (1 + log x)
For stationary point, we put f ‘(x) = 0
⇒ x^x (1 + log x) = 0
⇒ 1 + log x = 0 [∵ x^x > o]
⇒ log x = – 1
⇒ x = e^-1
= \(\frac{1}{e}\)
When x slightly < \(\frac{1}{e}\)
⇒ log x < – 1
⇒ 1 + log x < 0
⇒ f'(x) < 0 When x slightly > \(\frac{1}{e}\)
⇒ log x > – 1
⇒ 1 + log x > 0
⇒ f’(x) > 0
Thus f'(x) changes its sign from -ve to +ve.
∴ x = \(\frac{1}{e}\) be a point of minima.

Question 43.
The maximum value of \(\frac{\log x}{x}\) is
(a) e
(b) 2e
(c) \(\frac{1}{e}\)
(d) \(\frac{2}{e}\)
Solution:
(c) \(\frac{1}{e}\)

Given y = \(\frac{\log x}{x}\)

Thus y is maximise at x = e
∴ Maximum value of f(x) = f(e) = \(\frac{\log e}{e}=\frac{1}{e}\).

Question 44.
The minimum value of \(\frac{x}{\log _e x}\) is
(a) e
(b) 1/e
(c) 1
(d) none of these
Solution:
(a) e

Given f(x) = \(\frac{x}{\log _e x}\) is defined for x > 0
∴ f'(x) = \(\frac{\log x \cdot 1-x \times \frac{1}{x}}{(\log x)^2}\)
= \(\frac{\log x-1}{(\log x)^2}\)
For maxima / minima,
f’ (x) = 0
⇒ log x = 1
⇒ log x = log e
⇒ x = e
when x slightly < e
⇒ log x< 1
⇒ log x – 1 < 0
⇒ f'(x) < 0 when x slightly > e
⇒ log x > log e
⇒ log x – 1 > 0
⇒ f'(x) > 0
Thusf’ (x) changes its sign from negative to positive when x increases through e.
Thus x = e is a point of local minima and local min. value of f(x) = f(e) = \(\frac{e}{log e}\) = e.

Question 45.
Maximum slope of the curve y = – x³ + 3x² + 9x – 27 is:
(a) 0
(b) 12
(c) 16
(d) 32
Solution:
(b) 12

Given equation of curve be
y = – x³ + 3x² + 9x – 27
∴ slope of curve m = – 3x² + 6x + 9
we want to maximise m.
∴ \(\frac{d m}{d x}\) = – 6x + 6
for maximalminíma, we have \(\frac{d m}{d x}\) = 0
⇒ – 6x + 6 = 0
⇒ x = 1
and \(\frac{d^2 m}{d x^2}\) = – 6
Thus at x = 1;
\(\frac{d^2 m}{d x^2}\) = – 6 < 0
Hence m is maximise when x = 1.
∴ Max value of m = – 3 (1)² + 6 + 9 = 12

Question 46.
If f(x) = Zx³ – 21x² + 36x – 30, then
(a) f(x) has minimum at x = 1
(b) f(x) has maximum at x = 6
(c) f(x) has maximum at x = 1
(d) f(x) has no maxima or minima
Solution:
(c) f(x) has maximum at x = 1

Given f(x) = 2x³ – 21x² + 36x – 30
f’(x) = 6x² – 42x + 36
= 6 (x² – 7x +6)
= 6 (x – 1) (x – 6)
For maxima / minima, f’ (x) = 0
⇒ 6 (x -1) (x – 6)O = 0
⇒ x = 1, 6
∴ f”(x) = 6 (2x – 7)
Now f”(1) = 6 (2 – 7) = – 30 < 0
∴ f(x) is maximise when x = 1
and f”(6) = 6 (12 – 7) = 30 > 0
Thus f(x) is minimise at x = 6.

Question 47.
The least value of the function f(x) = ax + \(\frac{b}{x}\) (x > 0, a > 0, b > 0) is
(a) \(\sqrt{ab}\)
(b) 2 \(\sqrt{ab}\)
(c) ab
(d) 2ab
Solution:
(b) 2 \(\sqrt{ab}\)

Given f(x) = ax + \(\frac{b}{x}\)
∴ f'(x) = a – \(\frac{b}{x^{2}}\)
For critical points ; f'(x) = 0
⇒ a – \(\frac{b}{x^{2}}\) = 0
⇒ x² = \(\frac{b}{a}\)
⇒ x = \(\sqrt{\frac{b}{a}}\) [∵ x > 0]
[Since a > 0, b > 0
∴ \(\sqrt{\frac{b}{a}}\) > 0
∴ x ≠ \(\sqrt{\frac{b}{a}}\)]
∴ least value of f(x) = f (\(\sqrt{\frac{b}{a}}\))
= a \(\sqrt{\frac{b}{a}}\) + \(\frac{b}{\sqrt{\frac{b}{a}}}\)
= \(\sqrt{ab}\) + \(\sqrt{ab}\)
= 2 \(\sqrt{ab}\)